The heights for young females (age 18 - 29) arc approximately normal with mean 6

Question

The heights for young females (age 18 - 29) arc approximately normal with mean 66.4 inches, and standard deviation of 4.2 inches. 1.What percentage of young females are shorter than 67 inches tall? 2.What percentage of young females are within than 5 to 6 feet tall? 4.How tall a young female should be in order to fall in the bottom 25% of all young women heights? 5.If a random sample of 36 young females is selected, what is the probability that the sample mean height is less than 67 inches? 6.If a random sample of 100 young females is sclectcd, what is the probability that at least 50 of them arc shorter than 67 inches tall?

Explanation / Answer

Here we are given that The heights for young females are approximately normal with mean = 66.4

and sd = 4.2

1) Here we have to find P(X<=67).

> pnorm(67,mean=66.4,sd=4.2)
[1] 0.5567985

Therefore percentage = 0.5567985*100 = 55.67985%

2) In this part we have to calculate P(5 < X < 6).

1 feet = 12 inches

then 5 feet = 12*5 = 60 inches

6 feet = 12*6 = 72 inches.

We know that,

P(60<X<72) = P(X<=72) - P(X<=60)

> pnorm(72,mean=66.4,sd=4.2)
[1] 0.9087888
> pnorm(60,mean=66.4,sd=4.2)
[1] 0.06377815

P(60<X<72) = 0.9088 - 0.0638 = 0.8450

Percentage = 0.8450*100 = 84.50%

3) In this part we have to calculate P(X > 1.8 meters)

We know that,

1 meter = 39.3701 inches

therefore 1.8 meter = ?

? = 39.3701*1.8 = 70.8661

That is now we have to find P(X > 70.8661).

P(X > 70.8661) = 1 - P(X<=70.8661)

> pnorm(70.8661,mean=66.4,sd=4.2)
[1] 0.85619

P(X<=70.8661) = 0.8562

P(X > 70.8661) = 1 - 0.8562 = 0.1438

Percentage = 0.1438*100 = 14.38%

4) In this part we have to calculate X when probability is given.

probability = 25% = 0.25

> qnorm(0.25,mean=66.4,sd=4.2)
[1] 63.56714

5) n = sample size = 36

Here we have to find P(Xbar <67).

Here we know that the distribution of sample mean is also normal with mean = 66.4 and sd = 4.2/sqrt(36) = 0.7 .

> pnorm(67,mean=66.4,sd=0.7)
[1] 0.804317

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