The waist circum. of males 20-29 yr old is approx normally distributed, mean u-92.5cm standard deviation o=13.7cmThe waist circumference of males 20-29 is appx normally distributed, with mean...

The waist circum. of males 20-29 yr old is approx normally distributed, mean u-92.5cm standard deviation o=13.7cm

The waist circumference of males 20-29 is appx normally distributed, with mean u=92.5cm and standard deviation o=13.7cm.

a. use model to determine the proportion of 20-29 males whose waist circumference is less than 100cm

b.Probability that randomly selected 20-29 male has waist circumference between 80 and 100cm?

c. Determine the waist circumferences that represent the middle 90% of all waist circumferences for 20-29 yr. old males?

d. Determine the waist circumference for 20-20 yr old males that is at the 10th percentile. 

Asked on by jjtyme

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embizze | High School Teacher | (Level 1) Educator Emeritus

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Given `mu=92.5,sigma=13.7` ** It is important to note that we are given `sigma` , the standard deviation of the population**

(a) Find the proportion of males whose circumference is less than 100

This is the same as the probability of a randomly selected individual having a size less than 100 or `P(x<100)` .

Converting to a standard normal score we get `z=(100-92.5)/13.7~~0.55` (My calculator gives .5474452555)

So `P(x<100)=P(z<0.55)` . From a standard normal table we find the area to the left of z=0.55 to be .7088 (With no rounding and using a calcultor I get `~~.7079635953` )

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The proportion of males size less than 100 is approximately 71%

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(b) Find `P(80<x<100)`

We have already converted 100 to a standard z score: `z_(100)~~0.55`

Converting 80 we get `z_(80)=(80-92.5)/13.7~~-0.91`

Then `P(80<x<100)=P(-0.91<z<0.55)` From the standard normal table we get the area to the left of z=-0.91 to be .1814

The area to the left of z=100 we found to be .7088

The area between is .7088-.1814=.5204

** My calculator gives .5274**

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The probability that a randomly selected male is sized between 80 and 100 is approximately .5204

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(c) Find the scores that represent the middle 90%.

This means that 5% are above and 5% are below.

` `The z score associated with an area of .0500 is -1.64 **Some books/instructors will have you round to -1.65 as this is between values on the table; others might have you interpolate; still others will have you use a calculator in which case you will get -1.644853626**

To find the x value which gives a z score of -1.64 we use `x=zsigma+mu` (Just solve `z=(x-mu)/sigma` for x)

So `x=(-1.64)(13.7)+92.5=70.032` (70 inches is 1.64 standard deviations below the mean)

Simalarily we find the z score for .9500 to be 1.64

So `x=1.64(13.7)+92.5=114.968`

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The middle 90% of sizes is approximately 70<x<115 inches

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(d) Find the waist size associated with the 10th percentile:

The area to the left of z is .1000; using the table we find z=-1.28

Then `x=-1.28(13.7)+92.5=74.964`

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The size associated with the 10th percentile is approximately 75 inches

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Sources:

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