Mathematics, 25.06.2020 03:01 Crxymia

Adult male heights are a normal random variable with mean 69 inches and a standard deviation of 3 inches. Find the height, to the nearest inch, for which only 8 percent of adult males are taller (i. find the 92nd percentile)

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The height (corresponding to the $\\ 92^{nd}$ percentile) is (to the nearest inch) 73 inches (and, approximately, only 8% of adult males are taller than this height.)

Step-by-step explanation:

Roughly speaking, the $\\ 92^{nd}$ percentile is the x value (in the distribution) for which 92% of the observations in the [normal] distribution are below this x value, and 8% of the observations are above this x value.

To answer this question, we already know that:

Heights are a normal random variable, i.e, it follows a normal distribution.The mean for this distribution is $\\ \mu = 69$ inches.The standard deviation is $\\ \sigma = 3$ inches.

Strategy for solving the question

For solving this, we have to use here the following key concepts: z-scores, the cumulative standard normal distribution, and the cumulative standard normal table.

Z-scores

To find the $\\ 92^{nd}$ percentile, we can use z-scores or standardized values. A z-score is a value that tells us the distance in standard deviations units from the mean. When the z-score is positive, it means that the value is above the mean. A negative indicates that the z-score is below the mean. The formula to obtain a z-score is as follows:

$\\ z = \frac{x - \mu}{\sigma}$ [1]

Where

z is the z-score.x is the raw score. $\\ \mu$ is the mean. $\\ \sigma$ is the standard deviation.

Cumulative standard normal distribution and corresponding table

We still need to know the corresponding z-score, z, for the cumulative probability of 92%. For this, we have to consult the standard normal table, available on the Internet or in any Statistics books.

In this case, we look in the different columns of the standard normal table a probability value (exact or approximate) to 0.92 and then find the value for z that corresponds to this probability. The value for z is between 1.40 (0.91924) and 1.41 (0.92073).

Using z = 1.40 in [1], we have:

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

$\\ 1.40 = \frac{x - 69}{3}$

Then, solving for x:

Multiplying by 3 at each side of the equation:

$\\ 1.40 * 3 = x - 69$

Adding 69 at both sides of the equation:

$\\ (1.40 * 3) + 69 = x$

$\\ x = (1.40 * 3) + 69$

$\\ x = 4.20 + 69$

$\\ x = 73.20$

That is, the $\\ 92^{nd}$ percentile is 73.20 inches, and to the nearest inch, this percentile is 73 inches.

This result indicates that, approximately, 92% of the heights are below 73 inches, and only 8% of heights are taller than this height.

The shaded area in the graph below shows an area of 0.08076 (8.076%) for 73.20 inches.

Adult male heights are a normal random variable with mean 69 inches and a standard deviation of 3 in

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Number 5 is 50 degreesnumber 6 is 45 degreesnumber 7 is 60 degreesnumber 8 is 45 degrees
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answer: 5

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