Question

Next four questions are based on the following information. assuming that heights of college women are normally distributed with mean 64 inches and standard deviation 3.4 inches , answer the following questions.
a.what percent of women are taller than 64 inches?
b.what percent of women are between 60.6 inches and 67.4 inches?
c.what percent of women are between 57.2 and 70.8 inches?
d.what percent of women are between 57.2 and 67.4 inches?

Answer 1

Answer:

(a) The percentage of women who are taller than 64 inches is 50%.

(b) The percentage of  women whose heights are between 60.6 inches and 67.4 inches is 68.26%.

(c) The percentage of  women whose heights are between 57.2 inches and 70.8 inches is 95.44%.

(d) The percentage of  women whose heights are between 57.2 inches and 67.4 inches is 65.48%.

Step-by-step explanation:

Let X = heights of college women.

It is provided that the heights of college women (X) follows a Normal distribution.

The mean and standard deviation of the heights of college women are:

$\mu=64\ inches\\\sigma=3.4\ inches$

*Use the standard normal table for the probabilities.*

(a)

Compute the proportion of women who are taller than 64 inches as follows:

$P(X>64)=P(\frac{X-\mu}{\sigma}>\frac{64-64}{3.4})\\=P(Z>0)\\=1-P(Z$

The percentage is: 0.50 × 100 = 50%.

Thus, the percentage of women who are taller than 64 inches is 50%.

(b)

Compute the proportion of  women whose heights are between 60.6 inches and 67.4 inches as follows:

$P(60.6$

The percentage is: 0.6826 × 100 = 68.26%.

Thus, the percentage of  women whose heights are between 60.6 inches and 67.4 inches is 68.26%.

(c)

Compute the proportion of  women whose heights are between 57.2 inches and 70.8 inches as follows:

$P(57.2$

The percentage is: 0.9544 × 100 = 95.44%.

Thus, the percentage of  women whose heights are between 57.2 inches and 70.8 inches is 95.44%.

(d)

Compute the proportion of  women whose heights are between 57.2 inches and 67.4 inches as follows:

$P(57.2$

The percentage is: 0.6548 × 100 = 65.48%.

Thus, the percentage of  women whose heights are between 57.2 inches and 67.4 inches is 65.48%.

Answer 2

Answer:

(a) 50%

(b) 68.3%

(c) 95.45%

(d) 81.86%

Step-by-step explanation:

We are given that heights of college women are normally distributed with mean 64 inches and standard deviation 3.4 inches i.e.;

Mean, $\mu$ = 64 inches   and   Standard deviation, $\sigma$ = 3.4 inches

Also,  Z = $\frac{X -\mu}{\sigma}$ ~ N(0,1)

(a) Let X = percent of women

  P(X > 64) = P( $\frac{X -\mu}{\sigma}$ > $\frac{64 -64}{3.4}$ ) = P(Z > 0) = 50%

(b) P(60.6 < X < 67.4) = P(X < 67.4) - P(X <= 60.6)

    P(X < 67.4)   = P( $\frac{X -\mu}{\sigma}$ < $\frac{67.4 -64}{3.4}$ )   = P(Z < 1) = 0.84134

    P(X <= 60.6) = P( $\frac{X -\mu}{\sigma}$ <= $\frac{60.6 -64}{3.4}$ ) = P(Z <= -1) = 1 - P(Z < 1)

                                                                               = 1 - 0.84134 = 0.15866

    P(X < 67.4) - P(X <= 60.6) = 0.84134 - 0.15866 = 0.68268 = 68.3%

(c) P(57.2 < X < 70.8) = P(X < 70.8) - P(X <= 57.2)

    P(X < 70.8)   = P( $\frac{X -\mu}{\sigma}$ < $\frac{70.8 -64}{3.4}$ )   = P(Z < 2) = 0.97725

    P(X <= 57.2) = P( $\frac{X -\mu}{\sigma}$ <= $\frac{57.2 -64}{3.4}$ ) = P(Z <= -2) = 1 - P(Z < 2)

                                                                               = 1 - 0.97725 = 0.02275

    P(X < 70.8) - P(X <= 57.2) = 0.97725 - 0.02275 = 0.9545 = 95.45%

(d) P(57.2 < X < 67.4) = P(X < 67.4) - P(X <= 57.2)

    P(X < 67.4)   = P( $\frac{X -\mu}{\sigma}$ < $\frac{67.4 -64}{3.4}$ )   = P(Z < 1) = 0.84134

    P(X <= 57.2) = P( $\frac{X -\mu}{\sigma}$ <= $\frac{57.2-64}{3.4}$ ) = P(Z <= -2) = 1 - P(Z < 1)

                                                                               = 1 - 0.97725 = 0.02275

    P(X < 67.4) - P(X <= 57.2) = 0.84134 - 0.02275 = 0.81859 = 81.86%