Question

31. Assume that​ women's heights are normally distributed with a mean given by μ=62.6 in​, and a standard deviation given by σ=2.8 in.

a. If 1 woman is randomly​ selected, find the probability that her height is between 62.2 in and 63.2 in.The probability is approximately

​(b) If 49 women are randomly​ selected, find the probability that they have a mean height less than 63 in

.43. In a survey of 1345 people, 1029 people said they voted in a recent presidential election. Voting records show that 74​% of eligible voters actually did vote. Given that 74​% of eligible voters actually did​ vote,

(a) find the probability that among 1345 randomly selected​ voters, at least 1029 actually did vote.​ (b) What do the results from part​ (a) suggest?

​(a) P(X≥1029​)=

Answer 1

31.

(a)

Given that X follows a Normal distribution with Mean = u = 62.6 S.D= 0 =2.8 Note that z= (x-u)/o = (x-62.6)/2.8 ---(Eq.1) follows standard Normal.

When x = 62.2, from (Eq.1), z = -0.14 When x = 63.2, from (Eq.1), z = 0.21 P(62.2 <X<63.2)=P(-0.14 <Z<0.21)=P(Z <0.21) - P(Z <-0.14)=0.5832 -0.4443 = 0.1389 (ans)

(b)

Note that sample mean X follows a Normal distribution with Mean = u = 62.6 Standard deviation = 0/Vn=0.4 Note that z= (x-1)/(o/Vn) = (x-62.6)/0.4 ---(Eq.1) follows standard Normal. When 7 = 63, from (Eq.1), z = 1 PĂ<63)=P(Z<1)=0.8413 (ans)

(probability values are found by Normal table)

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