Question

Assuming that the heights of college women are normally distributed with mean 70 inches and standard deviation 3.4 inches, answer the following questions. (Hint: Use the figure below with mean μ and standard deviation σ.)

(a) What percentage of women are taller than 70 inches?
%

(b) What percentage of women are shorter than 70 inches?
%

(c) What percentage of women are between 66.6 inches and 73.4 inches?
%

(d) What percentage of women are between 63.2 and 76.8 inches?
%

Answer 1

solution:

the given data as follows:

mean = = 70

standard deviation = = 3.4

a) percentage of women taller than 70 inches

since it is given that the height are normally distributed. so in the normal distribution the shape of the curve is symmatric or bell shaped and mean devide the distribution in to two equal parts 50% area bove the mean and 50% area below the mean . the following graph describe the situation

SO the percentage of women above the mean =50%

b)

from the graph represented above we can say that

percentage of women below the mean = 50%

c) percentage of women between 66.6 and 73.4 inches

first of all we have to find the z score: z score describes the value how much standard deviation far from the mean

for X = 66.6

for X = 73.4,

so P(-1< z < 1) so we have to find the percentage of height fall within 1 standard deviation from the mean

the below graph shows the emperical rule of probability that 68% of the values fall within 1 standard deviatio from the mean

SO percentage of women are between 66.6 inch and 73.4 inch = 68%

d) percentage of women height between 63.2 and 76.8 inches

so these value describe the area which fall which fall within 2 standard deviation from the mean

according to emperical rule 95% of data fall within 2 standard deviation from the mean as shown in the above graph.

so percentage of women height between 63.2 and 76.8 inches = 95%