Question

In a survey of women in a certain country ( ages 20-29), the mean height was 65.6 inches with a standard deviation of 2.87 inches. Answer the following questions about the specified normal distribution. (a) What height represents the 99th percentile? (b) What height represtents the firsth quartile? (Round to two decimal places as needed)

Answer 1

Answer:

a) 72.28 inches

b) 63.67 inches

Step-by-step explanation:

Mean Height = u = 65.6 inches

Standard Deviation = $\sigma$ = 2.87 inches

The data is said to be Normally Distributed.

Part a)

We have to find the 99th percentile. Since the data is normally distributed, we can use the z table to find the desired answer.

99th percentile means 99% of the values are below this point, this gives a probability value of 0.99. From the z table we have to look for the z score with a probability corresponding to 0.99. This value comes out to be 2.326

i.e. P(z < 2.326) = 0.99

Since, now we have the z score we need to find the equivalent height using this z score. The formula of z score is:

$z=\frac{x-u}{\sigma}$

Using the values, we get:

$2.326=\frac{x-65.6}{2.87}\\\\ x = 2.87 \times 2.326 + 65.6\\\\ x = 72.28$

Thus, P ( x < 72.28 ) = 0.99

Therefore, 72.28 inches represent the 99th percentile of the data.

Part b)

We have to find the First Quartile. First Quartile means 25th percentile. Using the same procedure as followed in previous part, first we need to find the z score. From the z table, we get:

P(z < -0.674) = 0.25

So, a z-score of -0.674 represents the 25th percentile. Now we need to find the equivalent height from this z score. Using the values in formula of z score again, we get:

$-0.674=\frac{x-65.6}{2.87} \\\\ x=2.87 \times (-0.674) + 65.6\\\\ x = 63.67$

So, P(x < 63.67) = 0.25

Thus, a height of 63.67 inches represent the 25th Percentile or the First Quartile.