# frequency distribution  4 results

frequency distribution - frequency measurement of various outcomes

A random sample of 25 customers was chosen in CCP MiniMart between 3:00 and 4:00 PM on a Friday afte
A random sample of 25 customers was chosen in CCP MiniMart between 3:00 and 4:00 PM on a Friday afternoon. The frequency distribution below shows the distribution for checkout time (in minutes). Checkout Time (in minutes) | Frequency | Relative Frequency 1.0 - 1.9 | 2 | ? 2.0 - 2.9 | 8 | ? 3.0 - 3.9 | ? | ? 4.0 - 5.9 | 5 | ? Total | 25 | ? (a) Complete the frequency table with frequency and relative frequency. (b) What percentage of the checkout times was less than 3 minutes? (c)In what class interval must the median lie? Explain your answer. (d) Assume that the largest observation in this dataset is 5.8. Suppose this observation were incorrectly recorded as 8.5 instead of 5.8. Will the mean increase, decrease, or remain the same? Will the median increase, decrease or remain the same? Why? (a) [B]Checkout Time (in minutes) | Frequency | Relative Frequency 1.0 - 1.9 | 2 | 2/25 2.0 - 2.9 | 8 | 8/25 3.0 - 3.9 | 10 (since 25 - 5 + 8 + 2) = 10 | 10/25 4.0 - 5.9 | 5 | 5/25 Total | 25 | ?[/B] (b) (2 + 8)/25 = 10/25 = [B]40%[/B] c) [B]3.0 - 3.9[/B] since 2 + 8 + 10 + 5 = 25 and 13 is the middle value which occurs in the 3.0 - 3.9 interval (d) [B]Mean increases[/B] since it's a higher value than usual. Median would not change as the median is the most frequent distribution and assuming the 5.8 is only recorded once.

Basic Statistics
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

Frequency Distribution Table
Free Frequency Distribution Table Calculator - Determines the classes and frequency distribution using the 2 to k rule.

The first significant digit in any number must be 1, 2, 3, 4, 5, 6, 7, 8, or 9. It was discovered t
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.