Free 2 Asset Portfolio Calculator - Given a portfolio with 2 assets, this determines the expected return (mean), variance, and volatility (standard deviation) of the portfolio.

A compact disc is designed to last an average of 4 years with a standard deviation of 0.8 years. What is the probability that a CD will last less than 3 years? Using our [URL='http://www.mathcelebrity.com/probnormdist.php?xone=3&mean=4&stdev=0.8&n=1&pl=P%28X+%3C+Z%29']Z-score and Normal distribution calculator[/URL], we get: [B]0.10565[/B]

A financial analyst computed the ROI for all companies listed on the NYSE. She found that the mean of this distribution was 10% with standard deviation of 5%. She is interested in examining further those companies whose ROI is between 14% and 16% of the approximately 1,500 companies listed on the exchange, how many are of interest of her? First, use our [URL='http://www.mathcelebrity.com/zscore.php?z=p%280.14%3Cz%3C0.16%29&pl=Calculate+Probability']z-score calculator[/URL] to get P(0.14 < z < 0.16) = 0.007889 Divide that by 2 for two-tail test to get0.003944729 Use the NORMSINV(0.003944729) in Excel to get the Z value of 2.66 Therefore, the companies of interest are 2.66 * 1500 * 0.10 = [B]399[/B]

A fuel injection system is designed to last 18 years, with a standard deviation of 1.4 years. What is the probability that a fuel injection system will last less than 15 years? Using our [URL='https://www.mathcelebrity.com/probnormdist.php?xone=15&mean=18&stdev=14&n=1&pl=P%28X+%3C+Z%29']z-score calculator[/URL], we see that: P(X < 15) = [B]0.416834[/B]

A group of students at a school takes a history test. The distribution is normal with a mean of 25, and a standard deviation of 4. (a) Everyone who scores in the top 30% of the distribution gets a certificate. (b) The top 5% of the scores get to compete in a statewide history contest. What is the lowest score someone can get and still go onto compete with the rest of the state? (a) [URL='http://www.mathcelebrity.com/zcritical.php?a=0.70&pl=Calculate+Critical+Z+Value']Top 30% is 70% percentile[/URL] Inverse of normal distribution(0.7) = -0.5244005 Plug into z-score formula, -0.5244005 = (x - 25)/4 [B]x = 22.9024[/B] (b) [URL='http://www.mathcelebrity.com/zcritical.php?a=0.95&pl=Calculate+Critical+Z+Value']Top 5% is 95% percentile[/URL] Inverse of normal distribution(0.95) = 1.644853627 Plug into z-score formula, 1.644853627 = (x - 25)/4 [B]x = 31.57941451[/B]

A random sample of 100 light bulbs has a mean lifetime of 3000 hours. Assume that the population standard deviation of the lifetime is 500 hours. Construct a 95% confidence interval estimate of the mean lifetime. [B]2902 < u < 3098[/B] using our [URL='http://www.mathcelebrity.com/normconf.php?n=100&xbar=3000&stdev=500&conf=95&rdig=4&pl=Large+Sample']confidence interval for the mean calculator[/URL]

A random sample of 144 with a mean of 100 and a standard deviation of 70 is known from a population of 1,000. What is the 95% confidence interval for the unknown population? [URL='http://www.mathcelebrity.com/normconf.php?n=144&xbar=100&stdev=70&conf=95&rdig=4&pl=Large+Sample']Large Sample Confidence Interval Mean Test[/URL] [B]88.5667 < u < 111.4333[/B]

A random sample of 149 students has a test score average of 61 with a standard deviation of 10. Find the margin of error if the confidence level is 0.99. (Round answer to two decimal places) Using our [URL='https://www.mathcelebrity.com/normconf.php?n=149&xbar=61&stdev=10&conf=99&rdig=4&pl=Large+Sample']confidence interval of the mean calculator[/URL], we get [B]58.89 < u < 63.11[/B]

A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.22 hours, with a standard deviation of 2.31 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.44 hours, with a standard deviation of 1.74 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no children and adults with children left parenthesis mu 1 minus mu 2 right parenthesis (?1 - ?2). Using our confidence interval for [URL='http://www.mathcelebrity.com/meandiffconf.php?n1=+40&xbar1=+5.22&stdev1=+2.31&n2=+40&xbar2=+4.44&stdev2=1.74&conf=+90&pl=Mean+Diff+Conf.+Interval+%28Large+Sample%29']difference of means calculator[/URL], we get: [B]0.0278 < ?1 - ?2 < 1.5322[/B]

A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.22 hours, with a standard deviation of 2.31 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.29 hours, with a standard deviation of 1.58 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no children and adults with children (u1 - u2) What is the interpretation of this confidence interval? A. There is 90% confidence that the difference of the means is in the interval. Conclude that there is insufficient evidence of a significant difference in the number of leisure hours B. There is a 90% probability that the difference of the means is in the interval. Conclude that there is a significant difference in the number of leisure hours C. There is 90% confidence that the difference of the means is in the interval. Conclude that there is a significant difference in the number of leisure hours D. There is a 90% probability that the difference of the means is in the interval. Conclude that there is insufficient evidence of a significant difference in the number of leisure hours 0.2021 < u1 - u2 < 1.6579 using our [URL='http://www.mathcelebrity.com/meandiffconf.php?n1=+40&xbar1=+5.22&stdev1=2.31&n2=40&xbar2=4.29&stdev2=1.58&conf=+90&pl=Mean+Diff+Conf.+Interval+%28Large+Sample%29']difference of means confidence interval calculator[/URL] [B]Choice D There is a 90% probability that the difference of the means is in the interval. Conclude that there is insufficient evidence of a significant difference in the number of leisure hours[/B]

A random sample of n = 10 flash light batteries with a mean operating life X=5 hr. And a sample standard deviation S = 1 hr. is picked from a production line known to produce batteries with normally distributed operating lives. What's the 98% confidence interval for the unknown mean of the working life of the entire population of batteries? [URL='http://www.mathcelebrity.com/normconf.php?n=10&xbar=5&stdev=1&conf=98&rdig=4&pl=Small+Sample']Small Sample Confidence Interval for the Mean test[/URL] [B]4.1078 < u < 5.8922[/B]

A random sample of STAT200 weekly study times in hours is as follows: 2 15 15 18 30 Find the sample standard deviation. (Round the answer to two decimal places. Show all work.) [B]9.98[/B] using [URL='http://www.mathcelebrity.com/statbasic.php?num1=+2,15,15,18,30&num2=+0.2,0.4,0.6,0.8,0.9&pl=Number+Set+Basics']our standard deviation calculator[/URL]

A random variable X follows the uniform distribution with a lower limit of 670 and an upper limit a. Calculate the mean and standard deviation of this distribution. (Round intermediate calculation for standard deviation to 4 decimal places and final answer to 2 decimal places.) Using our [URL='http://www.mathcelebrity.com/uniform.php?a=+670&b=+770&x=+680&t=+3&pl=PDF']uniform distribution calculator[/URL], we get: [B]Mean = 720 Standard deviation = 28.87 [/B] b. What is the probability that X is less than 730? (Do not round intermediate calculations. Round your answer to 2 decimal places.) Using our [URL='http://www.mathcelebrity.com/uniform.php?a=+670&b=+770&x=+730&t=+3&pl=CDF']uniform distribution calculator[/URL], we get: [B]0.6[/B]

A researcher posed a null hypothesis that there was no significant difference between boys and girls on a standard memory test. He randomly sampled 100 girls and 120 boys in a community and gave them the standard memory test. The mean score for girls was 70 and the standard deviation of mean was 5.0. The mean score for boys was 65 and the standard deviation of mean was 6.0. What's the absolute value of the difference between means? |70 -65| = |5| = 5

A researcher posed a null hypothesis that there was no significant difference between boys and girls on a standard memory test. He randomly sampled 100 girls and 100 boys in a community and gave them the standard memory test. The mean score for girls was 70 and the standard deviation of mean was 5.0. The mean score for boys was 65 and the standard deviation of mean was 5.0. What is the standard error of the difference in means? [B]0.707106781187[/B] using our [URL='http://www.mathcelebrity.com/meandiffconf.php?n1=+100&xbar1=70&stdev1=5&n2=+100&xbar2=65&stdev2=5&conf=+99&pl=Hypothesis+Test']difference of means calculator[/URL]

A researcher posed a null hypothesis that there was no significant difference between boys and girls on a standard memory test. He randomly sampled 100 girls and 100 boys in a community and gave them the standard memory test. The mean score for girls was 70 and the standard deviation of mean was 5.0. The mean score for boys was 65 and the standard deviation of mean was 5.0. What's the t-value (two-tailed) for the null hypothesis that boys and girls have the same test scores? t = 7.07106781187 using our [URL='http://www.mathcelebrity.com/meandiffconf.php?n1=+100&xbar1=70&stdev1=5&n2=+100&xbar2=65&stdev2=5&conf=+99&pl=Hypothesis+Test']difference of means calculator[/URL]

A toy factory makes 5,000 teddy bears per day. The supervisor randomly selects 10 teddy bears from all 5,000 teddy bears and uses this sample to estimate the mean weight of teddy bears and the sample standard deviation. How many degrees of freedom are there in the estimate of the standard deviation? DF = n - 1 DF = 10 - 1 [B]DF = 9[/B]

A typical adult has an average IQ score of 105 with a standard deviation of 20. If 20 randomly selected adults are given an IQ test, what is the probability that the sample mean scores will be between 85 and 125 points? For x = 125, our z-score and probability is seen [URL='http://www.mathcelebrity.com/probnormdist.php?xone=+125&mean=+105&stdev=+20&n=+1&pl=P%28X+%3C+Z%29']here[/URL] Z = 1 P(x < 1) = 0.841345 For x = 85, our z-score and probability is seen [URL='http://www.mathcelebrity.com/probnormdist.php?xone=+85&mean=+105&stdev=+20&n=+1&pl=P%28X+%3C+Z%29']here[/URL] Z = -1 P(x < -1) = 0.158655 So what we want is the probability between these values:
0.841345 - 0.158655 = [B]0.68269[/B]

An analysis of the final test scores for Managerial Decision Making Tools reveals the scores follow the normal probability distribution. The mean of the distribution is 75 and the standard deviation is 8. The instructor wants to award an "A" to students whose score is in the highest 10 percent. What is the dividing point for those students who earn an "A"? Top 10% is equivalent to the 90th percentile. Using our [URL='http://www.mathcelebrity.com/percentile_normal.php?mean=+75&stdev=8&p=+90&pl=Calculate+Percentile']percentile calculator[/URL], the 90th percentile cutoff point is [B]85.256[/B]

As the sample size increases, we assume: a. ? increases b. ? increases c. The probability of rejecting a hypothesis increases d. Power increases [B]d. Power increases[/B] [LIST] [*]Power increases if the standard deviation is smaller. [*]If the difference between the means is bigger, the power is bigger. [*]Sample size also increases power [/LIST]

Assume the speed of vehicles along a stretch of I-10 has an approximately normal distribution with a mean of 71 mph and a standard deviation of 8 mph. a. The current speed limit is 65 mph. What is the proportion of vehicles less than or equal to the speed limit? b. What proportion of the vehicles would be going less than 50 mph? c. A new speed limit will be initiated such that approximately 10% of vehicles will be over the speed limit. What is the new speed limit based on this criterion? d. In what way do you think the actual distribution of speeds differs from a normal distribution? a. Using our [URL='http://www.mathcelebrity.com/probnormdist.php?xone=65&mean=71&stdev=8&n=+1&pl=P%28X+%3C+Z%29']z-score calculator[/URL], we see that P(x<65) = [B]22.66%[/B] b. Using our [URL='http://www.mathcelebrity.com/probnormdist.php?xone=+50&mean=71&stdev=8&n=+1&pl=P%28X+%3C+Z%29']z-score calculator[/URL], we see that P(x<50) = [B]0.4269%[/B] c. [URL='http://www.mathcelebrity.com/zcritical.php?a=0.9&pl=Calculate+Critical+Z+Value']Inverse of normal for 90% percentile[/URL] = 1.281551566 Plug into z-score formula: (x - 71)/8 = 1.281551566 [B]x = 81.25241252[/B] d. [B]The shape/ trail differ because the normal distribution is symmetric with relatively more values at the center. Where the actual has a flatter trail and could be expected to occur.[/B]

Did they give you a standard deviation?

Free Basic Statistics Calculator - Given a number set, and an optional probability set, this calculates the following statistical items:
Expected Value
Mean = μ
Variance = σ²
Standard Deviation = σ
Standard Error of the Mean
Skewness
Mid-Range
Average Deviation (Mean Absolute Deviation)
Median
Mode
Range
Pearsons Skewness Coefficients
Entropy
Upper Quartile (hinge) (75th Percentile)
Lower Quartile (hinge) (25th Percentile)
InnerQuartile Range
Inner Fences (Lower Inner Fence and Upper Inner Fence)
Outer Fences (Lower Outer Fence and Upper Outer Fence)
Suspect Outliers
Highly Suspect Outliers
Stem and Leaf Plot
Ranked Data Set
Central Tendency Items such as Harmonic Mean and Geometric Mean and Mid-Range
Root Mean Square
Weighted Average (Weighted Mean)
Frequency Distribution
Successive Ratio

Below are data showing the results of six subjects on a memory test. The three scores per subject are their scores on three trials (a, b, and c) of a memory task. Are the subjects getting better each trial? Test the linear effect of trial for the data. A score trial B score trial 2 C Score trial 3 4 6 7 3 7 8 2 8 5 1 4 7 4 6 9 2 4 2 (a) Compute L for each subject using the contrast weights -1, 0, and 1. That is, compute (-1)(a) + (0)(b) + (1)(c) for each subject. (b) Compute a one-sample t-test on this column (with the L values for each subject) you created. Formula t = To computer a one-sample t-test first know the meaning of each letter (a) Each L column value is just -1(Column 1) + 0(Column2) + 1(Column 3) A score trial B score trial 2 C Score trial 3 L = (-1)(a) + (0)(b) + (1)(c) 4 6 7 3 3 7 8 5 2 8 5 3 1 4 7 6 4 6 9 5 2 4 2 0 (b) Mean = (3 + 5 + 3 + 6 + 5 + 0)/6 = 22/6 = 3.666666667 Standard Deviation = 2.160246899 Use 3 as our test mean (3.666667 - 3)/(2.160246899/sqrt(6)) = 0.755928946

Free Binomial Distribution Calculator - Calculates the probability of 3 separate events that follow a binomial distribution. It calculates the probability of exactly k successes, no more than k successes, and greater than k successes as well as the mean, variance, standard deviation, skewness and kurtosis.
Also calculates the normal approximation to the binomial distribution with and without the continuity correction factor
Calculates moment number t using the moment generating function

Free Chebyshevs Theorem Calculator - Using Chebyshevs Theorem, this calculates the following:
Probability that random variable X is within k standard deviations of the mean.
How many k standard deviations within the mean given a P(X) value.

CHEBYSHEVS THEOREM TELLS US THAT WHAT PERCENTAGE LIES BETWEEN 2.25 STANDARD DEVIATIONS? Using our [URL='http://www.mathcelebrity.com/chebyshev.php?pl=probability&k=2.25&probk=0.75']Chebyshevs Theorem calculator[/URL], we get: P(X - u| < k?) >= [B]0.802469[/B]

Compute a 75% Chebyshev interval around the mean for [I]x[/I] values and also for [I]y[/I] values. [B][U]Grid E: [I]x[/I] variable[/U][/B] 11.92 34.86 26.72 24.50 38.93 8.59 29.31 23.39 24.13 30.05 21.54 35.97 7.48 35.97 [B][U]Grid H: [I]y[/I] variable[/U][/B] 27.86 13.29 33.03 44.31 16.58 42.43 39.61 25.51 39.14 16.58 47.13 14.70 57.47 34.44 According to Chebyshev's Theorem, [1 - (1/k^2)] proportion of values will fall between Mean +/- (k*SD) k in this case equal to z z = (X-Mean)/SD X = Mean + (z*SD) 1 - 1/k^2 = 0.75 - 1/k^2 = 0.75 - 1= - 0.25 1/k^2 = 0.25 k^2 = 1/0.25 k^2 = 4 k = 2 Therefore, z = k = 2 First, [URL='http://www.mathcelebrity.com/statbasic.php?num1=11.92%2C34.86%2C26.72%2C24.50%2C38.93%2C8.59%2C29.31%2C23.39%2C24.13%2C30.05%2C21.54%2C35.97%2C7.48%2C35.97&num2=+0.2%2C0.4%2C0.6%2C0.8%2C0.9&pl=Number+Set+Basics']determine the mean and standard deviation of x[/URL] Mean(x) = 25.24 SD(x) = 9.7873 Required Interval for x is: Mean - (z * SD) < X < Mean + (z * SD) 25.24 - (2 * 9.7873) < X < 25.24 - (2 * 9.7873) 25.24 - 19.5746 < X < 25.24 + 19.5746 5.6654 < X < 44.8146 Next, [URL='http://www.mathcelebrity.com/statbasic.php?num1=27.86%2C13.29%2C33.03%2C44.31%2C16.58%2C42.43%2C39.61%2C25.51%2C39.14%2C16.58%2C47.13%2C14.70%2C57.47%2C34.44&num2=+0.2%2C0.4%2C0.6%2C0.8%2C0.9&pl=Number+Set+Basics']determine the mean and standard deviation of y[/URL] Mean(y) = 32.29 SD(y) = 9.7873 Required Interval for y is: Mean - (z * SD) < Y < Mean + (z * SD) 32.29 - (2 * 13.1932) < Y < 32.29 - (2 * 13.1932) 32.29 - 26.3864 < Y < 32.29 + 26.3864 5.9036 < X < 58.6764

Free Confidence Interval for Variance and Standard Deviation Calculator - Calculates a (95% - 99%) estimation of confidence interval for the standard deviation or variance using the χ² method with (n - 1) degrees of freedom.

Free Exponential Distribution Calculator - Calculates the Probability Density Function (PDF) and Cumulative Density Function (CDF) of the exponential distribution as well as the mean, variance, standard deviation, and entropy.

Facebook provides a variety of statistics on its Web site that detail the growth and popularity of the site. On average, 28 percent of 18 to 34 year olds check their Facebook profiles before getting out of bed in the morning. Suppose this percentage follows a normal distribution with a standard deviation of five percent. a. Find the probability that the percent of 18 to 34-year-olds who check Facebook before getting out of bed in the morning is at least 30. b. Find the 95th percentile, and express it in a sentence. a. P(X >=0.30), calculate the [URL='http://www.mathcelebrity.com/probnormdist.php?xone=+0.30&mean=+0.28&stdev=+0.05&n=+1&pl=P%28X+%3E+Z%29']z-score[/URL] which is: Z = 0.4 P(x>0.4) = [B]0.344578 or 34.46%[/B] b. Inverse Normal (0.95) [URL='http://www.mathcelebrity.com/zcritical.php?a=0.95&pl=Calculate+Critical+Z+Value']calculator[/URL] = 1.644853627 Use NORMSINV(0.95) on Excel 0.28 + 0.05(1.644853627) = [B]0.362242681 or 36.22%[/B]

mean of 106 inches and a standard deviation of 10 inches and for sample of size is 25. Determine the mean and the standard deviation of /x

Do you mean x bar? mean of 106 inches and a standard deviation of 10 inches and for sample of size is 25. Determine the mean and the standard deviation of /x If so, x bar equals the population mean. So it's [B]106[/B]. Sample standard deviation = Population standard deviation / square root of n 10/Sqrt(25) 10/5 [B]2[/B]

one trunk can carry 5068.8 lb. weight of boxes that it carries have a mean of 75lb and a standard deviation of 16 Ib. For Sample size of 64 ,find the mean and standard deviation of /x

one trunk can carry 5068.8 lb. weight of boxes that it carries have a mean of 75lb and a standard deviation of 16 Ib. For Sample size of 64 ,find the mean and standard deviation of /x

Sample Mean = Population Mean Sample Mean = [B]75[/B] Sample Standard Deviation = Population Standard Deviation / Sqrt(n) Sample Standard Deviation = 16/sqrt(64) Sample Standard Deviation = 16 / 8 Sample Standard Deviation = [B]2[/B]

The monthly earnings of a group of business students are are normally distributed with a standard deviation of 545 dollars. A researcher wants to estimate the mean monthly earnings of all business students. Find the sample size needed to have a confidence level of 95% and a margin of error of 128 dollars.

Free Geometric Distribution Calculator - Using a geometric distribution, it calculates the probability of exactly k successes, no more than k successes, and greater than k successes as well as the mean, variance, standard deviation, skewness, and kurtosis.
Calculates moment number t using the moment generating function

Height and weight are two measurements used to track a child's development. The World Health Organization measures child development by comparing the weights of children who are the same height and the same gender. In 2009, weights for all 80 cm girls in the reference population had a mean μ = 10.2 kg and standard deviation σ = 0.8 kg. Weights are normally distributed. X ~ N(10.2, 0.8). Calculate the z-scores that correspond to the following weights and interpret them. a. 11 kg
b. 7.9 kg
c. 12.2 kg a. [URL='http://www.mathcelebrity.com/probnormdist.php?xone=+11&mean=10.2&stdev=8&n=+1&pl=1" target="_blank']Answer A[/URL] - Z = 0.1 b. [URL='http://www.mathcelebrity.com/probnormdist.php?xone=+7.9&mean=+10.2&stdev=+8&n=+1&pl=1']Answer B[/URL] - Z = -0.288 c. [URL='http://www.mathcelebrity.com/probnormdist.php?xone=+12.2&mean=+10.2&stdev=+8&n=+1&pl=1']Answer C[/URL] - Z = 0.25

A sample mean, sample size, and population standard deviation are given. Use the one-mean z-test to perform the required hypothesis test at the given significance level. x = 20.5, n = 11, ? = 7, H0: µ = 18.7 , Ha: µ ? 18.7 , ? = 0.01

sample mean, sample size, and population standard deviation are given. Use the one-mean z-test to perform the required hypothesis test at the given significance level. x = 3.7, n = 32, ? = 1.8, H0: µ = 4.2 , Ha: µ ? 4.2 , ? = 0.05

A sample mean, sample size, and population standard deviation are given. Use the one-mean z-test to perform the required hypothesis test about the mean, µ, of the population from which the sample was drawn x = 3.26 , S = 0.55, ?N= 9, H0: µ = 2.85, Ha: µ > 2.85 , ? = 0.01

If approximately 68% of the scores lie between 40.7 and 59.9 , then the approximate value of the standard deviation for the distribution, according to the empirical rule, is The empirical rule states 68% of the values lie within 1 standard deviation of the mean. The mean is the midpoint of the interval above: (59.9 + 40.7)/2 = 50.3 Standard deviation is the absolute value of the mean - endpoint |59.9 - 50.3| = [B]9.6[/B]

If the distribution of IQ scores is bell-shaped, with a mean of 100 and a standard deviation of 15, then approximately ____% of IQ scores are less than 55? A bell-shaped curved implies a normal distribution. By using our [URL='https://www.mathcelebrity.com/probnormdist.php?xone=55&mean=100&stdev=15&n=1&pl=Empirical+Rule']empirical rule calculator[/URL], we see that: 99.7% of all normal distribution values lie within 3 standard deviations of the mean. This means the percent of scores less than 55 which is 3 standard deviations away from the mean is: 100% - 99.7% = [B]0.3%[/B]

Imagine that a researcher wanted to know the average weight of 5th grade boys in a high school. He randomly sampled 5 boys from that high school. Their weights were: 120 lbs., 99 lbs, 101 lbs, 87 lbs, 140 lbs. What's the sample [U][B]standard deviation[/B][/U]? [B]20.79182532[/B] using stdev.s in excel or also found on our [URL='http://www.mathcelebrity.com/statbasic.php?num1=120%2C99%2C101%2C87%2C140&num2=+0.2%2C0.4%2C0.6%2C0.8%2C0.9&pl=Number+Set+Basics#standard_deviation']statistics calculator[/URL]

Imagine that a researcher wanted to know the average weight of 5th grade boys in a high school. He randomly sampled 5 boys from that high school. Their weights were: 120 lbs., 99 lbs, 101 lbs, 87 lbs, 140 lbs. The researcher posed a null hypothesis that the average weight for boys in that high school should be 100 lbs. What is the [B][U]absolute value[/U][/B] of calculated t that we use for testing the null hypothesis? Mean is 109.4 and Standard Deviation = 20.79182532 using our [URL='http://www.mathcelebrity.com/statbasic.php?num1=120%2C99%2C101%2C87%2C140&num2=+0.2%2C0.4%2C0.6%2C0.8%2C0.9&pl=Number+Set+Basics']statistics calculator[/URL] Now use those values and calculate the t-value Abs(t value) = (100 - 109.4)/ 20.79182532/sqrt(5) Abs(tvalue) = [B]1.010928029[/B]

In the wild, monkeys eat an average of 28 bananas a day with a standard deviation of 2 bananas. One monkey eats only 21 bananas. What is the z-score for this monkey? Is the number of bananas the monkey eats unusually low? Using [URL='https://www.mathcelebrity.com/probnormdist.php?xone=21&mean=28&stdev=2&n=1&pl=P%28X+%3C+Z%29']our z-score calculator[/URL], we get: Z < -3.5 P(Z < -3.5) = 0.499767 Also, this [B]is unusually low as it's more than 3 deviations away from the mean[/B]

Is 3 standard deviations above the means considered an outlier? [B]Yes.[/B] Using the empirical rule, we know that: [LIST] [*]68% of the values lie within one standard deviation of the mean [*]95% of the values lie within two standard deviations of the mean [/LIST] Anything out side of two standard deviations is considered an outlier.

It is estimated that weekly demand for gasoline at new station is normally distributed, with an average of 1,000 and standard deviation of 50 gallons. The station will be supplied with gasoline once a week. What must the capacity of its tank be if the probability that its supply will be exhausted in a week is to be no more than 0.01? 0.01 is the 99th percentile Using our [URL='http://www.mathcelebrity.com/percentile_normal.php?mean=+1000&stdev=50&p=99&pl=Calculate+Percentile']percentile calculator[/URL], we get [B]x = 1116.3[/B]

Men's heights are normally distributed with mean 69.0 inches and standard deviation 2.8 inches. Mimi is designing a plane with a height that allows 95% of the men to stand straight without bending in the plane. What is the minimum height of the plane? Using the [URL='http://www.mathcelebrity.com/probnormdist.php?xone=50&mean=69&stdev=2.8&n=1&pl=Empirical+Rule']empirical rule calculator[/URL], we have a [B]63.4[/B] minimum height.

Free Negative Binomial Distribution Calculator - Calculates the probability of the k^th success on the x^th try for a negative binomial distribution also known as the Pascal distribution.? ? It calculates the probability of exactly k successes, no more than k successes, and greater than k successes as well as the mean, variance, and standard deviation.

Free Percentile for Normal Distribution Calculator - Given a mean, standard deviation, and a percentile range, this will calculate the percentile value.

Free Poisson Distribution Calculator - Calculates the probability of 3 separate events that follow a poisson distribution.
It calculates the probability of exactly k successes P(x = k)
No more than k successes P (x <= k)
Greater than k successes P(x >= k)
Each scenario also calculates the mean, variance, standard deviation, skewness, and kurtosis.
Calculates moment number t using the moment generating function

Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test, [U][B]the Type I error is[/B][/U]: a. to conclude that the current mean hours per week is higher than 4.5, when in fact, it is higher
b. to conclude that the current mean hours per week is higher than 4.5, when in fact, it is the same
c. to conclude that the mean hours per week currently is 4.5, when in fact, it is higher
d. to conclude that the mean hours per week currently is no higher than 4.5, when in fact, it is not higher [B]b. to conclude that the current mean hours per week is higher than 4.5, when in fact, it is the same [/B] [I]A Type I error is when you reject the null hypothesis when it is in fact true[/I]

Imagine you are in a game show. Prizes hidden on a game board with 10 spaces. One prize is worth $100, another is worth $50, and two are worth $10. You have to pay $20 to the host if your choice is not correct. Let the random variable x be the winning (a) What is your expected winning in this game? (b) Determine the standard deviation of x. (Round the answer to two decimal places) (a) 100(0.1) + 50(0.1) + 10(0.2) - 20 = 10 + 5 + 2 - 20 = [B]-3[/B] (b) 3.3 using our [URL='http://www.mathcelebrity.com/statbasic.php?num1=+100,50,10&num2=+0.1,0.1,0.2&usep=usep&pl=Number+Set+Basics']standard deviation calculator[/URL]

Free Sample Size Reliability for μ Calculator - Given a population standard deviation σ, a reliability (confidence) value or percentage, and a variation, this will calculate the sample size necessary to make that test valid.

Free Sample Size Requirement for the Difference of Means Calculator - Given a population standard deviation 1 of σ₁, a population standard deviation 2 of σ₂ a reliability (confidence) value or percentage, and a variation, this will calculate the sample size necessary to make that test valid.

Free Sharpe Ratio Calculator - Calculates the Sharpe ratio given return on assets, risk free rate, and standard deviation

standard deviation of 545 dollars. Find the sample size needed to have a confidence level of 95% and margin of error 128

Standard Error (margin of Error) = Standard Deviation / sqrt(n) 128 = 545/sqrt(n) Cross multiply: 128sqrt(n) = 545 Divide by 128 sqrt(n) = 4.2578125 Square both sides: [B]n = 18.1289672852 But we need an integer, so the answer is 19[/B]

Suppose a firm producing light bulbs wants to know if it can claim that its light bulbs it produces last 1,000 burning hours (u). To do this, the firm takes a random sample of 100 bulbs and find its average life time (X=980 hrs) and the sample standard deviation s = 80 hrs. If the firm wants to conduct the test at the 1% of significance, what's you final suggestion? (i..e, Should the producer accept the Ho that its light bulbs have a 1,000 burning hrs. at the 1% level of significance?) Ho: u = 1,000 hours. Ha: u <> 1,000 hours. [URL='http://www.mathcelebrity.com/mean_hypothesis.php?xbar=+980&n=+100&stdev=+80&ptype==&mean=+1000&alpha=+0.01&pl=Mean+Hypothesis+Testing']Perform a hypothesis test of the mean[/URL] [B]Yes, accept null hypothesis[/B]

Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet. We randomly sample 49 fly balls. a. If X = average distance in feet for 49 fly balls, then X ~ _______(_______,_______)
b. What is the probability that the 49 balls traveled an average of less than 240 feet? Sketch the graph. Scale the horizontal axis for X. Shade the region corresponding to the probability. Find the probability.
c. Find the 80^th percentile of the distribution of the average of 49 fly balls a. N(250, 50/sqrt(49)) = [B]0.42074[/B] b. Calculate Z-score and probability = 0.08 shown [URL='http://www.mathcelebrity.com/probnormdist.php?xone=+240&mean=+250&stdev=+7.14&n=+1&pl=P%28X+%3C+Z%29']here[/URL] c. Inverse of normal distribution(0.8) = 0.8416. Use NORMSINV(0.8) [URL='http://www.mathcelebrity.com/zcritical.php?a=0.8&pl=Calculate+Critical+Z+Value']calculator[/URL] Using the Z-score formula, we have 0.8416 = (x - 250)/50 x = [B]292.08[/B]

Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet. a. If X = distance in feet for a fly ball, then X ~ b. If one fly ball is randomly chosen from this distribution, what is the probability that this ball traveled fewer than 220 feet? Sketch the graph. Scale the horizontal axis X. Shade the region corresponding to the probability. Find the probability. c. Find the 80th percentile of the distribution of fly balls. Sketch the graph, and write the probability statement. a. [B]N(250, 50/sqrt(1))[/B] b. Calculate [URL='http://www.mathcelebrity.com/probnormdist.php?xone=+220&mean=250&stdev=50&n=+1&pl=P%28X+%3C+Z%29']z-score[/URL] Z = -0.6 and P(Z < -0.6) = [B]0.274253[/B] c. Inverse of normal distribution(0.8) = 0.8416 using NORMSINV(0.8) [URL='http://www.mathcelebrity.com/zcritical.php?a=0.8&pl=Calculate+Critical+Z+Value']calculator[/URL] Z-score formula: 0.8416 = (x - 250)/50
x = [B]292.08[/B]

The average precipitation for the first 7 months of the year is 19.32 inches with a standard deviation of 2.4 inches. Assume that the average precipitation is normally distributed. a. What is the probability that a randomly selected year will have precipitation greater than 18 inches for the first 7 months? b. What is the average precipitation of 5 randomly selected years for the first 7 months? c. What is the probability of 5 randomly selected years will have an average precipitation greater than 18 inches for the first 7 months? [URL='https://www.mathcelebrity.com/probnormdist.php?xone=18&mean=19.32&stdev=2.4&n=1&pl=P%28X+%3E+Z%29']For a. we set up our z-score for[/URL]: P(X>18) = 0.7088 b. We assume the average precipitation of 5 [I]randomly[/I] selected years for the first 7 months is the population mean ? = 19.32 c. [URL='https://www.mathcelebrity.com/probnormdist.php?xone=18&mean=19.32&stdev=2.4&n=5&pl=P%28X+%3E+Z%29']P(X > 18 with n = 5)[/URL] = 0.8907

The hourly wages of employees at Rowan have a mean wage rate of $10 per hour with a standard deviation of $1.20. What is the probability the mean hourly wage of a random sample of 36 employees will be larger than $10.50? Assume the company has a total of 1,000 employees Using our [URL='http://www.mathcelebrity.com/probnormdist.php?xone=10.5&mean=10&stdev=1.2&n=36&pl=P%28X+>+Z%29']normal distribution calculator[/URL], we get P(x > 10.5) = [B]0.00621[/B]

The IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. a) What is the probability that a randomly person has an IQ between 85 and 115? b) Find the 90th percentile of the IQ distribution c) If a random sample of 100 people is selected, what is the standard deviation of the sample mean? a) [B]68%[/B] from the [URL='http://www.mathcelebrity.com/probnormdist.php?xone=50&mean=100&stdev=15&n=1&pl=Empirical+Rule']empirical rule calculator[/URL] b) P(z) = 0.90. so z = 1.28152 using Excel NORMSINV(0.9)
(X - 100)/10 = 1.21852 X = [B]113[/B] rounded up c) Sample standard deviation is the population standard deviation divided by the square root of the sample size 15/sqrt(100) = 15/10 =[B] 1.5[/B]

The monthly earnings of a group of business students are are normally distributed with a standard deviation of 545 dollars. A researcher wants to estimate the mean monthly earnings of all business students. Find the sample size needed to have a confidence level of 95% and a margin of error of 128 dollars.

The monthly earnings of a group of business students are are normally distributed with a standard deviation of 545 dollars. A researcher wants to estimate the mean monthly earnings of all business students. Find the sample size needed to have a confidence level of 95% and a margin of error of 128 dollars.

[URL]http://mathcelebrity.com/community/threads/standard-deviation-of-545-dollars-find-the-sample-size-needed-to-have-a-confidence-level-of-95-and.450/[/URL]

The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days. What is the median recovery time? a. 2.7 b. 5.3 c. 7.4 d. 2.1 [B]b. 5.3 (mean, median, and mode are all the same in a normal distribution)[/B]

First, determine the [URL='http://www.mathcelebrity.com/statbasic.php?num1=87.4%2C86.9%2C89.9%2C78.3%2C75.1%2C70.6&num2=+0.2%2C0.4%2C0.6%2C0.8%2C0.9&pl=Number+Set+Basics']mean and standard deviation[/URL] for the [I]sample[/I] Mean = 81.3667 SD = 7.803 Next, use our [URL='http://www.mathcelebrity.com/normconf.php?n=6&xbar=81.3667&stdev=7.803&conf=90&rdig=4&pl=Small+Sample']confidence interval for the mean calculator[/URL] with these values and n = 6 [B]74.9478 < u < 87.7856[/B]

The relief time provided by a standard dose of a popular children’s allergy medicine averages 7.9 hours with a standard deviation of 2.2 hours. Use Table 1. a. Determine the percentage of children who experience relief for less than 6.4 hours if the relief time follows a normal distribution. (Round your answer to 2 decimal places.) Using our [URL='http://www.mathcelebrity.com/probnormdist.php?xone=6.4&mean=7.9&stdev=2.2&n=1&pl=P%28X+%3C+Z%29']normal distribution calculator[/URL], we get Answer = [B]0.25[/B]

The weight of a 9.5-inch by 6-inch paperback book published by Leaden Publications, Inc., is 16.2 oz. The standard deviation is 2.9 oz. What is the probability that the average weight of a sample of 33 such books is less than 15.89 oz? Using our [URL='http://www.mathcelebrity.com/probnormdist.php?xone=15.89&mean=16.2&stdev=2.9&n=33&pl=P%28X+%3C+Z%29']z-score calculator[/URL], we get: [B]0.271[/B]

True or False: The standard deviation of the chi-square distribution is twice the mean. [B]False[/B], the variance is twice the mean. Mean is k, Variance is 2k

Free Uniform Distribution Calculator - This calculates the following items for a uniform distribution
* Probability Density Function (PDF) ƒ(x)
* Cumulative Distribution Function (CDF) F(x)
* Mean, Variance, and Standard Deviation
Calculates moment number t using the moment generating function

Which of the following can increase power? a. Increasing standard deviation b. Decreasing standard deviation c. Increasing both means but keeping the difference between the means constant d. Increasing both means but making the difference between the means smaller [B]b. Decreasing standard deviation[/B] [LIST=1] [*]Power increases if the standard deviation is smaller. [*]If the difference between the means is bigger, the power is bigger. [*]Sample size increase also increases power [/LIST]

Which of the following involves making pairwise comparisons? a. Comparing the standard deviation of GRE grades between two states b. Comparing the variance of the amount of soda consumed by boys and girls in a high school c. Comparing the mean weight between children in grades 2, 3, 4 and 5 d. Comparing the number of restaurants in New York and Boston [B]c. Comparing the mean weight between children in grades 2, 3, 4 and 5[/B] Pairwise comparison generally refers to any process of comparing entities in pairs to judge which of each entity is preferred, or has a greater amount of some quantitative property

Which of the followings can increase the value of t? (select all the apply) a. Increase the standard deviation of difference scores b. Decrease the standard deviation of difference scores c. Increase the difference between means d. Decrease the difference between means [B]b. Decrease the standard deviation of difference scores c. Increase the difference between means[/B] [I]Increase numerator or decrease denominator of the t-value formula[/I]