Question

Suppose a normal distribution has a mean of 62 and a standard deviation of

4. What is the probability that a data value is between 55 and 63? Round your
answer to the nearest tenth of a percent.
O A. 55.9%
O B. 56.9%
O C. 53.9%
O D. 54.9%

Answer 1

Answer:To solve this problem, we need to standardize the values of 55 and 63 to z-scores and then use the standard normal distribution table to find the area between those two z-scores.

The z-score for a value of 55 is:

z = (55 - 62) / 4 = -1.75

The z-score for a value of 63 is:

z = (63 - 62) / 4 = 0.25

Using a standard normal distribution table, we can find that the area to the left of z = -1.75 is 0.0401 and the area to the left of z = 0.25 is 0.5987. Therefore, the area between z = -1.75 and z = 0.25 is:

0.5987 - 0.0401 = 0.5586

To convert this area back to a percentage, we multiply by 100:

0.5586 * 100 ≈ 55.9%

Therefore, the answer is A. 55.9%.

Explanation: