Suppose a sample of O-rings was obtained and the wall thickness (in inches) of e

Question

Suppose a sample of O-rings was obtained and the wall thickness (in inches) of each was recorded. Use a normal probability plot to assess whether the sample data could have come from a population that is normally distributed 0.187 0.196 0.188 0.200 0.204 0.218 0.233 0.236 0.256 0.270 0.278 0.291 0.297 0.307 0.309 0.318 Using the correlation coefficient of the normal probability plot, is it reasonable to conclude that the population is normally distributed? Select the correct choice below and fill in the answer boxe within your choice (Round to three decimal places as needed.) A. Yes. The correlation between the expected z-scores and the observed data. , exceeds the critical value. I. Therefore, it is reasonable to conclude that the data come from a norma 0 B. No. The correlation between the expected z-scores and the observed data. . does not exceed the critical value. I. Therefore, it is not reasonable to conclude that the data come O C. No. The correlation between the expected z-scores and the observed data, does not exceed the critical value, herfor, it is reasonable to conclude that the data come from D Yes. The correlation between the expected z scores and he observed data , exceeds he critical value Therefore it is not reasonable to conclude that the data come from a population. from a normal population. normal population. normal population. 1 Table of critical values Sample Size, n Critical Value Sample Size,n Crit 0.880 0.888 0.898 0.906 0.912 0.918 0.923 0.928 0.932 0.935 0.939 0.941 0.944 0.946 0.949 0.951 0.952 0.954 0.956 0.957 0.959 0.960 16 17 18 19 20 6 9 10 23 24 25 30 12 13 Click to select and enter your answer(s) 15

Explanation / Answer

a.
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 50
To Find standard deviation Value
P ( Z < X ) = 0.75
Value of z to the cumulative probability of 0.75 from normal table is 0.6745
P( X-u/s.d < X - 50/ s.d ) = 0.75
That is, ( 60 - 50/ s.d ) = 0.6745
--> s.d = ( 60 - 50 ) / 0.6745
--> s.d = ( 10 ) / 0.67
--> s.d = 14.93
b.
P ( Z < X ) = 0.6
Value of z to the cumulative probability of 0.6 from normal table is 0.2533
P( X-u/s.d < X - U /10 ) = 0.6
That is, ( 75 - U /10 ) = 0.2533
--> ( 75 - U ) = 0.2533*10
--> ( 75 - U /10 ) = 2.5335
--> U = -2.5335 + 75 = 72.46654

the mean must be lessthan 75

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