In a country, the mean time served by prisoners released from prison for the fir
ID: 2945731 • Letter: I
Question
In a country, the mean time served by prisoners released from prison for the first time is 32.3 months. Assume the standard deviation of the times served is 16.8 months. Let x denote time served to first release by a prisoner. Complete parts (a) through (e) below. a. Find the standardized version of x. (Type an expression using x as the variable. Do not simplify.) b. Find the mean and standard deviation of the standardized variable The mean of z isand the standard deviation of z is c. Determine the z-scores for prison times served of 76.2 months and 19.2 months. Round your answers to two decimal places. The z-score for a prison time served of 76.2 months is z (Round to two decimal places as needed.) The z-score for a prison time served of 19.2 months is z2- (Round to two decimal places as needed.) d. Interpret your answers in part (c) Interpret the z-score for a prison time served of 76.2 months. The z-score for a prison time served of 76.2 months isstandard deviationsthe mean prison time (Round to two decimal places as needed.) Interpret the z-score for a prison time served 19.2 months. The z-score for a prison time served of 19.2 months is standard deviations | ?| the mean prison time (Round to two decimal places as needed.) e. Construct a graph that depicts your results from parts (b) and (?). Denote 76.2 as x1 and 19.2 as x. Choose the correct answer beow Click to select your answer(s).Explanation / Answer
a)
z =(x-32.3)/16.8
b)
the mean of z is 0 and the std deviation of z is 1
c)
z score for a prison time served of 76.2 months is z1=(76.2-32.3)/16.8=2.61
z score for a prison time served of 19.2 months is z2=-0.78
d)
z score for a prison time served of 76.2 months is 2.61 std deviation above the mean prison time
z score for a prison time served of 19.2 months is 0.78 std deviation below the mean prison time