Question 4 of 25
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 58 and 64? Round your
answer to the nearest tenth of a percent.
A. 53.3%
B. 54.3%
C. 52.3%
D. 51.3%

Respuesta :

Answer:

Step-by-step explanation:

We are looking for P(58 < x < 64). We need to find the percentage to the left of the z-scores for each of these numbers. To find the z scores, use the formula:

[tex]z=\frac{x_i-\bar{x}}{\sigma}[/tex]

[tex]z=\frac{58-62}{4}[/tex] which gives us a z-score of -1. The percentage of numbers to the left of a z-score of -1 is .1586553

Now for the other z-score:

[tex]z=\frac{64-64}{4}[/tex] which gives us a z-score of .5. The percentage of numbers to the left of a z-score of .5 is .69146246

The lower percentage subtracted from the higher gives the area in question:

.69146246 - .1586553 = .53280716, or as a percentage, 53.3%, choice A.