Question

A sample of men's heights was taken and the mean was 68.8 inches. The standard deviation is 2.8 inches. What percent of the men in the sample were greater than 72 inches?

Answer 1

Answer:

87.29%

Step-by-step explanation:

Given: Mean= 68.8 inches

           Standard deviation= 2.8 inches

           

Now, finding the percent of the men in the sample were greater than 72 inches.

We know, z-score= $\frac{x-mean}{standard\ deviation}$

z-score= $\frac{72-68.8}{2.8}$

⇒ z-score= $\frac{3.2}{2.8} = 1.14$

∴ z-score= 1.14

Next, using normal distribution table to find percentage.

Coverting 0.8729 into percentage= $0.8729\times 100$

We get the percentage as 87.29%

Hence, 87.29% of the men in the sample were greater than 72 inches.