Given that women have normally distributed heights with a mean of 62.5 in. and a

ID: 3290817 • Letter: G

Question

Given that women have normally distributed heights with a mean of 62.5 in. and a standard deviation of 3.7 in., find the following:

The probability that a randomly selected woman has a height less than 65 in. In other

words find

(

)

The probability that a randomly se

lected woman has height between 60 in. and 63 in. In

other words find

(

)

Find the 75

percentile,

. In other words, find the height that separates the bottom

75% from the top 25%.

Show your work.

Use the range rule of thumb to identify the ma

ximum and minimum

usual heights. The formula for range rule of thumb can be fo

We take the help of Z tables to solve the problem:

Params of normal distribution have been given:

mean = 62.5

stdev = 3.7

a.P(X<65) = P(Z< (65-62.5)/ 3.7)= P(Z<.676) = .7517

b.P(60<X<63) = P(-.67<Z<.135) = .5557-.2483 = .3084

c.P( X<c) =.75, Z=.675, c=.675*3.7+62.5= 65

d. we know the thimb rule .

s R / 4

R = Maximum - Minimum

Where,

s = Standard Deviation

R = Range

So, range = 3.7*4 = 14.8

Thats' how to calculate the range, but there isn't a way to calculate the Maximum and minimum here

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