Assume the heights of women in a population follow the normal distribution with mean 64.3 and standard deviation 2.6 in.

a. a woman is randomly selected from the population. calculate the probability that the height for the woman is:

i. between 60 and 66 inches tall.

ii. over 67 inches tall

b. how tall is a woman whose height is at the 90th percentile?

c. what is the median height of women?

d. a certain woman has a height that is 0.5 standard deviations above the mean. what proportion of women are taller than she?

e. what height separates the smallest 10% of heights from the largest 90%?

f. five women are chosen at random from this population. what is the probability exactly one of them is more than 67 inches tall?

Step-by-step explanation:

given that the  heights of women in a population follow the normal distribution with mean 64.3 and standard deviation 2.6 in.

Let X be the height of the woman

a)

i)

ii)

b) 90th percentile is 67.628

(Mean + 1.28 times std deviation)

c) Median = mean sinc enormal distribution

Median = 64.3 inches

d) P(Z>0.5) =

e) 10th percentile =60.972

Difference between 90th and 10th percentile

=  

e) Each women is independent of the other.

For any woman to be taller than 67 inches is 0.1495 is constant for each women

So Y no of women taller than 67 is binomial with n =5

P(Y=1)= 0.3911

step-by-step explanation:

no, her answer is not correct.   we square the lengths of all sides and then determine whether the pythagorean theorem holds here:

9² + 15² = 306.   this is not equal to the square of the longest side ( that square being 395.   answer c is correct.

srry ask agian

step-by-step explanation: