6.3.23-T Question Hop 6.3.23-T A survey found that women\'s heights ane normally

Question

6.3.23-T Question Hop 6.3.23-T A survey found that women's heights ane normally distributed with mean 633 in, and standard deviation 25 in. The survey also found that men's heights are nomaly distributed with a mean 685in and standard deviation 2.8 Complete parts a through c below. a. Most of the ive characters at an amusement park have height of 4ft9in, and a maemum of 6ft4 in. Find me percentage of women meeting height requirement Stacrunch The percentage of women who meet the height requirement is (Round to two decimal places as needed.) Pearson Tutor Enter your answer in the answer box and then click Check Answer.

Explanation / Answer

mean = 63.3 and SD = 2.5 for women

we need to find 4feet 9 and 6 feet4

so 4 feet 9 = 4.9/12 = 4.75*12=57 and 6feet4 = 6.33*12= 76inch

P(57<X<76) , lets convert this into a z score so that we can check the probability values in Z table , please keep them handy

Use formula = (X-Mean)/SD = (57-63.3)/2.5 = -2.52

(76-63.3)/2.5 = 5.08

so now we must check P(-2.52 <Z<5.08)

To find the probability of P (2.52<Z<5.08), we use the following formula:

P (2.52<Z<5.08 )=P ( Z<5.08 )P (Z<2.52 )

We see that P ( Z<2.52 )=0.9941 so,

At the end we have:

Hope this helps !!

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