Question

Assume that the distribution of weights of adult men in the United States is normal with mean 190 pounds and standard deviation 30 pounds. The weight of a randomly selected adult male in the United States marks the start of the 40th percentile. How much does he weigh? Enter a number rounded to two decimal places, e.g. 165.12 not 165.12876. Do not enter the units.

Answer 1

Answer:

182.41

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 190, \sigma = 30$

40th percentile

Value of X when Z has a pvalue of 0.4. So X when Z = -0.253.

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

$-0.253 = \frac{X - 190}{30}$

$X - 190 = -0.253*30$

$X = 182.41$

So the answer is 182.41.