Question

The figure below shows a frequency and​ relative-frequency distribution for the heights of female students attending a college. Records show that the mean height of these students is

64.564.5

inches and that the standard deviation is

1.81.8

inches. Use the given information to complete parts​ (a) through​ (c).

Height​ (in.)	Frequency f	Relative freq.
60dash–under 61	22	0.00680.0068
61dash–under 62	66	0.02050.0205
62dash–under 63	2929	0.09900.0990
63dash–under 64	6565	0.22180.2218
64dash–under 65	9191	0.31060.3106
65dash–under 66	6969	0.23550.2355
66dash–under 67	2222	0.07510.0751
67dash–under 68	66	0.02050.0205
68dash–under 69	33	0.01020.0102
	293293	1.0000

a. The area under the normal curve with parameters

mu equals 64.5μ=64.5

and

sigma equals 1.8σ=1.8

that lies to the left of

6464

is

0.39470.3947.

Use this information to estimate the percent of female students who are shorter than

6464

inches.

nothing​%

​(Type an integer or a decimal. Do not​ round.)

Answer 1

We have given = 64.5 , standard deviation = 1.8

And we have also given

The area under the normal curve with parameters μ=64.5 and sigma σ=1.8 that lies to the left of 64 is 0.3947

That is P(x < 64) = 0.3947

And now we have asked to find the percent of female students who are shorter than 64 inches that is x<64

We have given that P( x < 64) = 0.3947 by multiplying this by 100 we get the percentage of female students who are shorter than 64.

Hence the percentage of female students who are shorter than 64 is= 0.3947 * 100 = 39.47 %

the percentage of female students who are shorter than 64 is  39.47 %

Hope this will help you. Thank you :)