Discussion on Frequency Distributions

Module 3 - Case
Frequency Distributions
Case Assignment
1 The final exam scores listed below are from one section of MATH 200.  How many scores were within one standard deviation of the mean? How many scores were within two standard deviations of the mean?
 99   34   86   57   73   85    91   93   46    96   88   79    68   85   89
Solution
First calculate the mean = (99+34+86+57+73+85+91+93+46+96+88+79+68+85+89)/15 = (1169/15)
Mean =77.93
Variance = ((99-77.93)2+(34-77.93)2+(86-77.93)2+(57-77.93)2+(73-77.93)2+(85-77.93)2+(91-77.93)2+(93-77.93)2+(46-77.93)2+(96-77.93)2+(88-77.93)2+(79-77.93)2+(68-77.93)2+(85-77.93)2+(89-77.93)2)/15
Variance = (443.94+1929.84+65.12+438.06+24.30+49.98+170.82+227.10+1019.52+326.52+101.40+1.14+98.60+49.98+122.54)/15
Variance =5068.86/15 =337.92
SD =√variance =√337.92 =18.38
Adding one SD to the mean =77.93+18.38=96.31
Subtracting one SD from the mean =77.93-18.38 =59.55
Scores that fall within 96.31 and 59.55 are 86, 73, 85, 91, 93, 96, 88 79, 68, 85, and 89.
Therefore, scores that were within one standard deviation of the mean are 11
Two SD =18.38×2 = 36.76
Adding two SD to mean = 77.93+36.76=114.69
Subtracting two SD from mean =77.93-36.76 =41.17
Scores that fall within 41.17 and 114.69 = 99, 86, 57, 73, 85, 91, 93 46, 96, 88, 79, 68, 85, and 89.
Therefore, the scores that were within two standard deviations of the mean were 14
1 The scores for math test #3 were normally distributed. If 15 students had a mean score of 74.8% and a standard deviation of 7.57, how many students scored above an 85%?
Solution
The Z score = (score given –mean)/SD
Z = (85-74.8)/7.57 =1.347
The probability associated with 1.347 = 0.9111
P (Students scored above 85) =1-0.9111
=0.0889
Of the 15 students (0.0889×15) scored 85% = 1 student
2 If you know the standard deviation, how do you find the variance?
Variance is obtained by finding the square of a known standard deviation.
That is;
Variance = (SD)2
3 To get the best deal on a stereo system, Louis called 8 out of 20 appliance stores in his neighborhood and asked for the cost of a specific model. The prices he was quoted are listed below:
       $216   $135   $281   $189   $218   $193   $299   $235
Find the standard deviation.
SD =√variance
Mean =(216+135+281+189+218+193+299+235)/8 =1766/8 =220.75
Variance = ((216-220.75)2+(135-220.75)2+(281-220.75)2+(189-220.75)2+(218-220.75)2+(193-220.75)2+(299-220.75)2+(235-220.75)2)/(8-1) since it is a sample
Variance = (22.56+7353.06+3630.06+1008.06+7.56+770.06+6123.06+203.06)/7= (19117.48/7) =2731.07
SD =52.26
1 A company has 70 employees whose salaries are summarized in the frequency distribution below.
Salary

Number of Employees

5,001–10,000

8

10,001–15,000

12
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