Assume that women's heights are normally distributed with a mean given by mu equals 62.5 in, and a standard deviation given by sigma equals 2.5 in. (a) If 1 woman is randomly selected, find the probability that her height is less than 63 in. (b) If 35 women are randomly selected, find the probability that they have a mean height less than 63 in. (a) The probability is approximately nothing. (Round to four decimal places as needed.) (b) The probability is approximately nothing. (Round to four decimal places as needed.)

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JeanaShupp JeanaShupp · Answer 1 · 2019-07-15T06:26:46+02:00

Answer: a) The probability is approximately = 0.5793

b) The probability is approximately=0.8810

Step-by-step explanation:

Given : Mean : [tex]\mu= 62.5\text{ in}[/tex]

Standard deviation : [tex]\sigma = \text{2.5 in}[/tex]

a) The formula for z -score :

[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]

Sample size = 1

For x= 63 in. ,

[tex]z=\dfrac{63-62.5}{\dfrac{2.5}{\sqrt{1}}}=0.2[/tex]

The p-value = [tex]P(z<0.2)=[/tex]

[tex]0.5792597\approx0.5793[/tex]

Thus, the probability is approximately = 0.5793

b) Sample size = 35

For x= 63 ,

[tex]z=\dfrac{63-62.5}{\dfrac{2.5}{\sqrt{35}}}\approx1.18[/tex]

The p-value = [tex]P(z<1.18)[/tex]

[tex]= 0.8809999\approx0.8810[/tex]

Thus , the probability is approximately=0.8810.