Question

Suppose the scores on a statistic exam are normally distributed with a mean of 77 and a variance of 25.

A. What is the 25th percentile of the scores?

B. What is the percentile of someone who got a score of 62?

C. What proportion of the scores are between 80 and 90?

D. Suppose you select 35 tests at random, what is the proportion of scores above 85?

Answer #1

Mean = 77

Variance = 25

Standard deviation = = 5

P(X < A) = P(Z < (A - mean)/standard deviation)

A. Let the 25th percentile be T

P(X < T) = 0.25

P(Z < (T - 77)/5) = 0.25

Take Z corresponding to 0.25 from standard normal distribution table

(T - 77)/5 = -0.67

T = **73.65**

B. P(X < 62) = P(Z < (62 - 77)/5)

= P(Z < -3)

= **0.0013**

C. P(80 < X < 90) = P(X < 90) - P(X < 80)

= P(Z < (90 - 77)/5) - P(Z < (80 - 77)/5)

= P(Z < 2.6) - P(Z < 0.6)

= 0.9953 - 0.7257

= **0.2696**

D. P(X > 85) = 1 - P(X < 85)

= 1 - P(Z < (85 - 77)/5)

= 1 - P(Z < 1.6)

= 1 - 0.9452

= **0.0548**

Number of scores out of 35 that are above 85 = 35x0.0548 = 1.918

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