52.3 is what percent of 1046
1 answer:
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Using the normal distribution, it is found that:
a. A male sailor whose waist is 34.1 inches is at the 87.5th percentile.
b. 5.7% of male sailors requires custom uniform pants.
<h3>Normal Probability Distribution</h3>The z-score of a measure X of a normally distributed variable with mean and standard deviation
is given by:
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
The mean and the standard deviation are given, respectively, by:
For item a, the percentile is the <u>p-value of Z when X = 34.1</u>, hence:
Z = (34.1 - 32.6)/1.3
Z = 1.15
Z = 1.15 has a p-value of 0.875.
Hence 87.5th percentile.
For item b, the proportion who does not require an special order is the <u>p-value of Z when X = 36 subtracted by the p-value of Z when X = 30,</u> hence:
X = 36:
Z = (36 - 32.6)/1.3
Z = 2.62
Z = 2.62 has a p-value of 0.996.
X = 30:
Z = (30 - 32.6)/1.3
Z = -2
Z = -2 has a p-value of 0.023.
0.996 - 0.023 = 0.943.
Hence the proportion who requires an special order is:
1 - 0.943 = 0.057.
5.7% of male sailors requires custom uniform pants.
More can be learned about the normal distribution at brainly.com/question/15181104
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