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]

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]

Free Critical Z-values Calculator - Given a probability from a normal distribution, this will generate the z-score critical value. Uses the NORMSINV Excel function.

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]

Free Random Sampling from the Normal Distribution Calculator - This performs hypothesis testing on a sample mean with critical value on a sample mean or calculates a probability that Z <= z or Z >= z using a random sample from a normal distribution.

Free Standard Normal Distribution Calculator - Givena normal distribution z-score critical value, this will generate the probability. Uses the NORMSDIST Excel function.

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 first significant digit in any number must be 1, 2, 3, 4, 5, 6, 7, 8, or 9. It was discovered that first digits do not occur with equal frequency. Probabilities of occurrence to the first digit in a number are shown in the accompanying table. The probability distribution is now known as Benford's Law. For example, the following distribution represents the first digits in 231 allegedly fraudulent checks written to a bogus company by an employee attempting to embezzle funds from his employer. Digit, Probability 1, 0.301 2, 0.176 3, 0.125 4, 0.097 5, 0.079 6, 0.067 7, 0.058 8, 0.051 9, 0.046 [B][U]Fradulent Checks[/U][/B] Digit, Frequency 1, 36 2, 32 3, 45 4, 20 5, 24 6, 36 7, 15 8, 16 9, 7 Complete parts (a) and (b). (a) Using the level of significance α = 0.05, test whether the first digits in the allegedly fraudulent checks obey Benford's Law. Do the first digits obey the Benford's Law?
Yes or No Based on the results of part (a), could one think that the employe is guilty of embezzlement? Yes or No Show frequency percentages Digit Fraud Probability Benford Probability 1 0.156 0.301 2 0.139 0.176 3 0.195 0.125 4 0.087 0.097 5 0.104 0.079 6 0.156 0.067 7 0.065 0.058 8 0.069 0.051 9 0.03 0.046 Take the difference between the 2 values, divide it by the Benford's Value. Sum up the squares to get the Test Stat of 2.725281277 Critical Value Excel: =CHIINV(0.95,8) = 2.733 Since test stat is less than critical value, we cannot reject, so [B]YES[/B], it does obey Benford's Law and [B]NO[/B], there is not enough evidence to suggest the employee is guilty of embezzlement.