Problem PageQuestion

Suppose that the heights of adult women in the United States are normally distributed with a mean of 65 inches and a standard deviation of 2.4 inches. Jennifer is taller than 70% of the population of U. S. women. How tall (in inches) is Jennifer? Carry your intermediate computations to at least four decimal places. Round your answer to one decimal place.

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:

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:

Thus, Jennifer's height is 63.7 inches.

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step-by-step explanation: