Question

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Answer 1

Answer:

Jennifer's height is 63.7 inches.

Step-by-step explanation:

Let X = heights of adult women in the United States.

The random variable X is normally distributed with mean, μ = 65 inches and standard deviation σ = 2.4 inches.

To compute the probability of a normal random variable we first need to convert the raw score to a standardized score or z-score.

The standardized score of a raw score X is:

$z=\frac{X-\mu}{\sigma}$

These standardized scores follows a normal distribution with mean 0 and variance 1.

It is provided that Jennifer is taller than 70% of the population of U.S. women.

Let Jennifer's height be denoted by x.

Then according to the information given:

    P (X > x) = 0.70

1 - P (X < x) = 0.70

    P (X < x) = 0.30

⇒  P (Z < z) = 0.30

The z-score related to the probability above is:

z = -0.5244

*Use a z-table.

Compute the value of x as follows:

          $z=\frac{X-\mu}{\sigma}$

$-0.5244=\frac{x-65}{2.4}$

          $x=65-(0.5244\times 2.4)$

             $=63.7414\\\approx63.7$

Thus, Jennifer's height is 63.7 inches.