Question

According to Bureau of Labor Statistics, 22.1% of the total part-time workforce in the U.S. was between the ages of 25 and 34 during the 3 rd quarter of 2011. A random sample of 80 part-time employees was selected during this quarter. Using the normal approximation to the binomial distribution, what is the probability that fewer than 20 people from this sample were between the ages of 25 and 34?

Answer 1

The probability is
`P(X < 20 ) = 0.68807`

Explanation:

From the question we are told that

The proportion of total part-time workforce is
`\r p = 0.221`

The sample size is n = 80

Generally the mean is mathematically represented as

`\mu = n* p`

`\mu = 0.221 * 80`

`\mu = 17.68`

The proportion of not part - time workforce

`q = 1- p`

=>
`q = 1- 0.221`

=>
`q = 0.779`

The standard deviation is mathematically represented as

`\sigma = √( 80 * 0.221 * 0.779 )`

`\sigma = 3.711`

Now applying the normal approximation,

Then the probability that fewer than 20 people from this sample were between the ages of 25 and 34 is mathematically represented as

`P(X < 20 ) = P( (X - \mu )/( \sigma ) < ( 20 - 17.68 )/( 3.711) )`

Applying continuity correction

`P(X < 20 ) = P( (X - \mu )/( \sigma ) < ( (20-0.5 ) - 17.68 )/( 3.711) )`

`P(X < 20 ) = P( (X - \mu )/( \sigma ) < ( (20-0.5 ) - 17.68 )/( 3.711) )`

`P(X < 20 ) = P( (X - \mu )/( \sigma ) < 0.4904 )`

Generally

`(X - \mu )/( \sigma ) = Z ( The \ standardized \ value \ of \ X )`

So

`P(X < 20 ) = P( Z< 0.4904 )`

From the z-table

`P( Z< 0.4904 ) = 0.68807`

The probability is

`P(X < 20 ) = 0.68807`