[Show all your work!] (1) Assume that you have a normal distribution of prior ar

Question

[Show all your work!] (1) Assume that you have a normal distribution of prior arrests with a mean of 7 and standard deviation of 3.09 Prior Arrests for A sample of 10 people. Prison Sentences (in months) for A sample of 10 people. 9 Person Number of Prior Arrests Z-score a. Calculate Z-scores for each person in the table b. What proportion of cases have fewer than 3 prior arrests? c. What proportion of cases have fewer than 5 prior arrests? d. What proportion of cases have more than 7 prior arrests? e. What proportion of cases have fewer than 7 prior arrests? f What proportion of cases have more than 8 prior arrests? g. What proportion of cases have more than 10 prior arrests? h. What proportion of cases have between 6 and 8 prior arrests? i. What proportion of cases have between 7 and 10 prior arrests? j What proportion of cases have between 5 and 7 prior arrests? k. What proportion of cases have between 8 and 10 prior arrests?

Explanation / Answer

Person

Number of Prior Arrests

Z score

1

2

-1.618

2

3

-1.294

3

5

-0.647

4

6

-0.323

5

6

-0.323

6

8

0.323

7

9

0.647

8

10

0.970

9

10

0.970

10

11

1.294

Z score = sample mean - population mean / standard deviation

Population mean = 7, Standard Deviation = 3.09

a. Z scores are given in the table above

using the Standard normal table we can calculate the proportion for below mentioned questions

b. proportion of cases have fewer than 3 prior arrests = P( Z -1.294) = 0.09 = 9%

c. Proportion of cases have fewer than 5 prior arrests = P( Z - 0.647) = 0.26 = 26%

d. proportion of cases have more than 7 arrest

Z score for each 7 = 7-7/3.09 = 0/3.09 = 0 = P( Z 0) = 0.5 = 50%

e. Proportion of cases have fewer than 7 prior arrest = P( Z 0) = 0.5 = 50%

f. proportion of cases have more than 8 prior arrests = P( Z 0.323) = 0.37 = 37%

g. proportion of cases have more than 10 prior arrest = P( Z 0.970) 0.16 = 16%

h proportion of cases have between 6 and 8 prior arrest

P (-0.323 < Z 0.323) = 0.12 + 0.12 = 0.24 = 24%

i. proportion of cases have between 7 and 10 prior arrests

P (0 <Z <0.970) = 0.33 = 33%

j. proportion of cases have between 5 and 7 prior arrests

P(-0.647 Z 0) = 0.23 = 23%

k proportion of cases have between 8 and 10 prior arrests

P(0.322 Z0.97) = 0.12 + 0.33 = 0.45 = 45%

Person

Number of Prior Arrests

Z score

1

2

-1.618

2

3

-1.294

3

5

-0.647

4

6

-0.323

5

6

-0.323

6

8

0.323

7

9

0.647

8

10

0.970

9

10

0.970

10

11

1.294

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