Question

of Alrican w Untitled docume Women on the Brea Women on the Discussion OutC Search Textbook https://webwork.uoregon.edu/webwork2/Math243Winter Bro Search arted Amaron.com-Online. Piceline.comProduct Support Dell Suggested Stes Web Slice Gl At least one of the answers above is NOT correct (1 point) The times for the mle run of a large group of male college students are approximately Normal with mean 7.2 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 midde 95% of this distributer? b) What percent of these men run a mle in more than 5.04 minutes? (a) from minutes bo Note: You can eam partial credt on this problem Preview My Answers Submit Answers Your score was recorded You have atempted this problem 3 Smes reconved a score of 0% lor ths atempt. Your overall recorded score is 0% You have unlimited attempts remaining to search 5 6 7 8 9

Answer 1

mean=7.2

sd=0.72

a) For 95% conf interval z value is +/-1.96 (From z table for area under the curve 2.5% towards each side)

Xbar = mean + z*sd

=7.2 -1.96*0.72

=5.7888

&

Xbar = mean + z*sd

=7.2 +1.96*0.72

=8.6112

So the 95% conf interval will range between 5.7888 to 8.6112

b) To find % people run more than 5.04 min

Z= (Xbar-mu)/sd

=(5.04-7.2)/0.72

=-3 from z table check for the area under the curve from -inf to -3 and subtract that from 1 so that we get the entire region of z>-3 we get the area = 1-0.0015 =0.9985

hence we can say that 99.85% is the population which run more than 5.04min

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