Introduction to Statistics Jordan Blausey 11/18/17 2:03 PM Homework: HW 8 ab Sco
ID: 3363714 • Letter: I
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Introduction to Statistics Jordan Blausey 11/18/17 2:03 PM Homework: HW 8 ab Score: 0 of 1 pt 80117(17 complete) HW Score: 72.7%, 12.36 of 17 pts rses10.1.17 Question Help * The frequency distribution shows the results of 200 test scores Are the test scores normally distributed? Use 005 Complete parts (a) theough (d Class boundaries T6 5-85 5 49.5-58 5 19 58.5-675 60 67 5-765 84 85 5-94.5 hypotheses are as follows Ho The test scores have a normal distrbution H, The test scores do not have a normal distribution read) & Tab a. Find the expected frequencies Calcu Frequency. Expected frequency ch Round to the nearest integer as needed.) Incon ork X Sorry, thatr's not correct At least one of your answers is incorrect. First, find the mean and standard deviation of the frequency distribution. Then use the mean and standard deviation to compute the z-score for each class boundary Then use z-scores to calculate the area under the standard normal curve for each class. Multiplying the resulting class areas by the sample size yields the expected frequency for each dass ok ia L OK an (o Enter your answer in the edit fields and then click Check Answer. Check Answer 4 Peamaining Clear AllExplanation / Answer
Here mean and variance calculation
So if the data is normally distributed than mean = 69.435
and standard deviation = sqrt (68.345775) = 8.2672
Here to find the cumlative probability we shall use the function.
Like for the interval 67.5 to 76.5 we find Z - value for both
Pr(67.5 < X < 76.5 ; 69.435; 8.267) would be calculated.
So expected values are
19, 63, 79 , 34 and 5 . Pleae put it into the table.
(b) CHi - square table also i will calculate.
p - value = Pr(X2 > 0.6887 ; df = 5-1 = 4) = 0.9462 > 0.05
so we cannot reject the null hypotheiss and can conclude that data is normally distributed.
Class Boundary Mid Value(x) Frequency (f) xf f*(x-x0)^2 49.5-58.5 54 19 1026 4526.5453 58.5-67.5 63 60 3780 2484.5535 67.5 - 76.5 72 84 6048 552.6549 76.5 -85.5 81 33 2673 4413.7244 85.5-94.5 90 4 360 1691.6769 Sum 200 13887 13669.155 Mean Variance 69.435 68.345775Related Questions
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