Question

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

Answer 1

Answer:

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) $P(60$

ii) $P(X>67) = 1-0.8505\\\\=0.1495$

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) = $1-0.6915\\=0.3085$

e) 10th percentile =60.972

Difference between 90th and 10th percentile

=  $67.628-60.972\\=6.656$

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