Question

question 17 of 25
suppose a normal distribution has a mean of 62 and a standard deviation of

1. what is the probability that a data value is between 54 and 66? round your

answer to the nearest tenth of a percent.

a. 78.9%
b. 79.9%
c. 77.9%
d. 81.9%

Explanation:

Step1: Calculate z-scores for 54 and 66

$z_1 = \frac{54 - 62}{4} = -2$, $z_2 = \frac{66 - 62}{4} = 1$

Step2: Find cumulative probabilities for z-scores

$P(Z < -2) ≈ 0.0228$, $P(Z < 1) ≈ 0.8413$

Step3: Compute the difference of probabilities

$P(-2 < Z < 1) = 0.8413 - 0.0228 = 0.8185$

Step4: Convert to percentage and round

$0.8185 × 100 ≈ 81.9\%$

Answer:

D. 81.9%