Question

Engineers must consider the breadths of male heads when designing helmets. The company researchers have determined that the population of potential clientele have head breadths that are normally distributed with a mean of 5.9-in and a standard deviation of 1.1-in.In what range would you expect to find the middle 98% of most head breadths?

Answer 1

Answer:

You should expect to find the middle 98% of most head breadths between 3.34 in and 8.46 in.

Step-by-step explanation:

When the distribution is normal, we use 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 question:

$\mu = 5.9, \sigma = 1.1$

In what range would you expect to find the middle 98% of most head breadths?

From the: 50 - (98/2) = 1st percentile.

To the: 50 + (98/2) = 99th percentile.

1st percentile:

X when Z has a pvalue of 0.01. So X when Z = -2.327.

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

$-2.327 = \frac{X - 5.9}{1.1}$

$X - 5.9 = -2.327*1.1$

$X = 3.34$

99th percentile:

X when Z has a pvalue of 0.99. So X when Z = 2.327.

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

$2.327 = \frac{X - 5.9}{1.1}$

$X - 5.9 = 2.327*1.1$

$X = 8.46$

You should expect to find the middle 98% of most head breadths between 3.34 in and 8.46 in.