Respuesta :
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:
[tex]\mu=64\ inches\\\sigma=3.4\ inches[/tex]
*Use the standard normal table for the probabilities.*
(a)
Compute the proportion of women who are taller than 64 inches as follows:
[tex]P(X>64)=P(\frac{X-\mu}{\sigma}>\frac{64-64}{3.4})\\=P(Z>0)\\=1-P(Z<0)\\=1-0.50\\=0.50[/tex]
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:
[tex]P(60.6<X<67.4)=P(X<67.4)-P(X<60.6)\\=P(\frac{X-\mu}{\sigma}<\frac{67.4-64}{3.4})-P(\frac{X-\mu}{\sigma}<\frac{60.6-64}{3.4})\\=P(Z<1)-P(Z<-1)\\=P(Z<1)-[1-P(Z<1)]\\=2P(Z<1)-1\\=(2\times0.8413)-1\\=0.6826[/tex]
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:
[tex]P(57.2<X<70.8)=P(X<70.8)-P(X<57.2)\\=P(\frac{X-\mu}{\sigma}<\frac{70.8-64}{3.4})-P(\frac{X-\mu}{\sigma}<\frac{57.2-64}{3.4})\\=P(Z<2)-P(Z<-2)\\=P(Z<2)-[1-P(Z<2)]\\=2P(Z<2)-1\\=(2\times0.9772)-1\\=0.9544[/tex]
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:
[tex]P(57.2<X<67.4)=P(X<67.4)-P(X<57.2)\\=P(\frac{X-\mu}{\sigma}<\frac{67.4-64}{3.4})-P(\frac{X-\mu}{\sigma}<\frac{57.2-64}{3.4})\\=P(Z<1)-P(Z<-2)\\=P(Z<1)-[1-P(Z<2)]\\=P(Z<1)+P(Z<2)-1\\=0.6826+0.9722-1\\=0.6548[/tex]
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:
(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, [tex]\mu[/tex] = 64 inches and Standard deviation, [tex]\sigma[/tex] = 3.4 inches
Also, Z = [tex]\frac{X -\mu}{\sigma}[/tex] ~ N(0,1)
(a) Let X = percent of women
P(X > 64) = P( [tex]\frac{X -\mu}{\sigma}[/tex] > [tex]\frac{64 -64}{3.4}[/tex] ) = P(Z > 0) = 50%
(b) P(60.6 < X < 67.4) = P(X < 67.4) - P(X <= 60.6)
P(X < 67.4) = P( [tex]\frac{X -\mu}{\sigma}[/tex] < [tex]\frac{67.4 -64}{3.4}[/tex] ) = P(Z < 1) = 0.84134
P(X <= 60.6) = P( [tex]\frac{X -\mu}{\sigma}[/tex] <= [tex]\frac{60.6 -64}{3.4}[/tex] ) = 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( [tex]\frac{X -\mu}{\sigma}[/tex] < [tex]\frac{70.8 -64}{3.4}[/tex] ) = P(Z < 2) = 0.97725
P(X <= 57.2) = P( [tex]\frac{X -\mu}{\sigma}[/tex] <= [tex]\frac{57.2 -64}{3.4}[/tex] ) = 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( [tex]\frac{X -\mu}{\sigma}[/tex] < [tex]\frac{67.4 -64}{3.4}[/tex] ) = P(Z < 1) = 0.84134
P(X <= 57.2) = P( [tex]\frac{X -\mu}{\sigma}[/tex] <= [tex]\frac{57.2-64}{3.4}[/tex] ) = 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%
