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Question

Suppose that an airline uses a seat width of 16.5in. Assume men have hip widths that are normally distributed with a mean of 14.4in. and a standard deviation of 1in. Find the probability that if an individual man is randomly​ selected, his hip width will be greater than 16.5in. ​(Round to four decimal places as​ needed.)


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Solution

Step-1: Find z score:

Given mean is μ is14.4in

Standard deviation, σ is 1in.

Let X be the hip width.

The distribution in the given question is normal distribution.

Therefore, standard normal distribution table z-table is a table used to know the percentage of values that are below z score in the standard normal distribution.

zscore=X-μσ1

where X is given value, μ is mean and σ is the standard deviation.

Substitute X=16.5, μ=14.4 and σ=1 in equation 1

zscore=16.5-14.41zscore=2.11zscore=2.1

Step-2: Find probability:

Probability that if an individual man is randomly​ selected, his hip width will be greater than 16.5in is

PX>16.5=Pz>zscorePX>16.5=Pz>2.1zscore=2.1PX>16.5=1-Pz2.1Pz>a=1-PzaPX>16.5=1-0.9821Fromstandardnormalz-table,Pz2.1=0.9821PX>16.5=0.0179

Hence, probability that if an individual man is randomly​ selected, his hip width will be greater than 16.5in is 0.0179in.


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