Assume that women\'s heights are normally distributed with a mean given by mu eq

Question

Assume that women's heights are normally distributed with a mean given by mu equals 63.6 in=63.6 in, and a standard deviation given by sigma equals 2.1 in=2.1 in. Complete parts a and b.

a. If 1 woman is randomly selected, find the probability that her height is between 63.4in and 64.4in.

Please show formulas, and math used.

Assume that women's heights are normally distributed with a mean given by H 63.6 in, and a standard deviation given by o 2.1 in. Complete parts a and b a. If 1 woman is randomly selected, find the probability that her height is between 63.4 in and 64.4 in. The probability is approximately (Round to four decimal places as needed.)

Explanation / Answer

Solution

Back-up Theory

If a random variable X ~ N(µ, 2), i.e., X has Normal Distribution with mean µ and variance 2, then

Z = (X - µ)/ ~ N(0, 1), i.e., Standard Normal Distribution ………………………..(1)

P(X or t) = P[{(X - µ)/ } or {(t - µ)/ }] = P[Z or {(t - µ)/ }] .………(2)

Probability values for the Standard Normal Variable, Z, can be directly read off from

Standard Normal Tables or using Excel Function……………………………………..(5

Now, to work out solution,

Let X = women’s height. Then, we are given X ~ N(63.6, 2.12).

We want P(63.4 X 64.4).

Now, [vide (2) under Back-up Theory],

P(63.4 X 64.4) = P[{(63.4 – 63.6)/2.1} Z {(64.4 – 63.6)/2.1}]

= P(- 0.0952 Z 0.3810) = P(Z 0.3810) - P(Z - 0.0952)

= 0.6483 – 0.4620 = 0.1863 ANSWER

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