Question

(1 point) The times for the mile run of a large group of male college students are approximately Normal with mean 7.02 minutes and standard deviation 0.72 minutes. Use the 68-95-99.7 rule to answer the following questions. (Start by making a sketch of the density curve you can use to mark areas on.) (a) what range of times covers the middle 99.7% of this distribution? (b) What percent of these men run a mile in less than 8.46 minutes? (a) from minutes to minutes

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Answer 1

a)accordang to central limit theorem the middle of 99.7 % covers 3 standard deviations.

Hence from 7.02-0.72*3 to 7.02+0.72*3

= 4.86 to 9.18

b) at 8.46 the Standard deviation count is 8.46-7.02=1.44 /0.72=2

Since at 2 standard deviation in an normal distribution it covers 97.5 %.

Next part

Since the distribution of women height follows a normal distribution with mean =64 and standard deviation as 2.67 , hence as asked the what percentage of women follows more height than men mean height.

Mean height of men is 69.3 so bg using. Z statistics

$Z= \frac{X-\mu}{\sigma}$

Z at X= 69.3 is 1.985 calculated hence or 2 hence here according to central limit theorem 2.357% of women height are more than mean men height.

Again,

a) If SAT distribution follow a normal distribution with mean =533 and standard deviation as 116 hence by using Z statistic formula

$Z= \frac{X-\mu}{\sigma}$

Z=1.612 now calculating the are more than at Z =1.612 by Z table or by calculator as 0.0534 means 5.34% age of men having scores 720 or more than 720.

b. Again if women SAT scores follows normal distribution hence again by Z statistic formula

$Z= \frac{X-\mu}{\sigma}$

Z at 720 calculated as 2.009 or 2 hence by central limit theorem 2.227 % of women have 720 or better.