In a survey of women in a certain country (ages 20 minus29), the mean height was 66.2 inches with a standard deviation of 2.86 inches. Answer the following questions about the specified normal distribution. (a) What height represents the 95th percentile? (round to two decimal places as needed) (b) What height represents the first quartile? (round to two decimal places as needed.)

(a) The height of 70.90 inches represents the 95th percentile.

(b) The height of 64.27 inches represents the first quartile.

Step-by-step explanation:

We are given that in a survey of women in a certain country (ages 20-29), the mean height was 66.2 inches with a standard deviation of 2.86 inches.

Let X = heights of women in a certain country.

So, X ~ Normal()

The z-score probability distribution for the normal distribution is given by;

                                   Z  =    ~ N(0,1)

where, = mean height = 66.2 inches

            = standard deviation = 2.86 inches

(a) We have to find the height that represents 95th percentile, that means;

         P(X < x) = 0.95        {wherex is the required height}

         P( < ) = 0.95

         P(Z <   ) = 0.95

Now, in the z table the critical value of x that represents the top 5% area is given by 1.645, that is;

                              x = 66.2 + 4.70 = 70.90 inches

So, the height of 70.90 inches represents the 95th percentile.

(b) We have to find the height that represents the first quartile, that means;

         P(X < x) = 0.25        {wherex is the required height}

         P( < ) = 0.25

         P(Z <   ) = 0.25

Now, in the z table the critical value of x that represents the below 25% area is given by -0.6745, that is;

                              x = 66.2 - 1.93 = 64.27 inches

So, the height of 64.27 inches represents the first quartile.

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