11. In a population of normally distributed aptitude scores of 1000 academy stud

ID: 3362085 • Letter: 1

Question

11. In a population of normally distributed aptitude scores of 1000 academy students with mean 70 and

Standard deviation 10, how many scores above 95?

a. 494 b. 6 c. 25 d. 250 e. not given

12. The z-score for x-bar = 2.7, mu = 3, s = 0.1, and n = 100 is:

a. –30 b. –3 c. –0.3 d. 3 e. not given

13. Find a 95% confidence interval for mu if s = 5.26, x-bar = 70.1, and n = 49.

a. 70.1+/-0.2104 b. 70.1+/-1.2361 c. 70.1+/-1.4728 d. 70.1+/-1.6772 e. not given

11)

mean i.e = 70

Standard deviation i.e = 10

z value = (x-)/

z value for 95 = (95-70)/10 = 25/10 = 2.5

p value for z = 2.5 using z table is 0.994 i.e P(score <= 95) = 0.994

P(score > 95) = 1- P(score <= 95) = 1 -0.994 = 0.006

so no of scores which are above 95 = 0.006*1000 = 6(b)

12)

= 3 ,s = 0.1 and n = 100

x = 2.7

standard error i.e s= 0.1

z score = (2.7 - 3)/0.1 = -0.3/0.1 = -30 (a)

13)

s = 5.26, x = 70.1, and n = 49

95% confidence interval is (mean - 1.96*sd, mean +1.96*sd)

standard deviation i.e = s/n = 5.26/49 = 5.26/7 = 0.7514

95% confidence interval for is (70.1-1.96*0.7514, 70.1+1.96*0.7514) i.e (70.1-1.4728 , 70.1+1.4728)

i.e 70.1 ± 1.4728 (c)

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