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?

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 and standard deviation , the zscore of a measure X is given by:

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:

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.

99th percentile:

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

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

show us the graphs dude

step-by-step explanation: