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lions [1.4K]
3 years ago
11

Assume that​ women's heights are normally distributed with a mean given by mu equals 62.3 in​, and a standard deviation given by

sigma equals 2.4 in.​(a) If 1 woman is randomly​ selected, find the probability that her height is less than 63 in.​(b) If 47 women are randomly​ selected, find the probability that they have a mean height less than 63 in.
Mathematics
1 answer:
Salsk061 [2.6K]3 years ago
8 0

Answer: a)  0.6141

b) 0.9772

Step-by-step explanation:

Given : Mean : \mu= 62.3\text{ in}

Standard deviation : \sigma = \text{2.4 in}

The formula for z -score :

z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}

a) Sample size = 1

For x= 63 in. ,

z=\dfrac{63-62.3}{\dfrac{2.4}{\sqrt{1}}}=0.29

The p-value = P(z

0.6140918\approx0.6141

Thus, the probability is approximately = 0.6141

b)  Sample size = 47

For x= 63 ,

z=\dfrac{63-62.3}{\dfrac{2.4}{\sqrt{47}}}\approx2.0

The p-value = P(z

=0.9772498\approx0.9772

Thus , the probability that they have a mean height less than 63 in =0.9772.

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