Answer:
The probability is [tex]P(X < 20 ) = 0.68807[/tex]
Step-by-step explanation:
From the question we are told that
The proportion of total part-time workforce is [tex]\r p = 0.221[/tex]
The sample size is n = 80
Generally the mean is mathematically represented as
[tex]\mu = n* p[/tex]
[tex]\mu = 0.221 * 80[/tex]
[tex]\mu = 17.68[/tex]
The proportion of not part - time workforce
[tex]q = 1- p[/tex]
=> [tex]q = 1- 0.221[/tex]
=> [tex]q = 0.779[/tex]
The standard deviation is mathematically represented as
[tex]\sigma = \sqrt{ 80 * 0.221 * 0.779 }[/tex]
[tex]\sigma = 3.711[/tex]
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
[tex]P(X < 20 ) = P( \frac{X - \mu }{ \sigma } < \frac{ 20 - 17.68 }{ 3.711} )[/tex]
Applying continuity correction
[tex]P(X < 20 ) = P( \frac{X - \mu }{ \sigma } < \frac{ (20-0.5 ) - 17.68 }{ 3.711} )[/tex]
[tex]P(X < 20 ) = P( \frac{X - \mu }{ \sigma } < \frac{ (20-0.5 ) - 17.68 }{ 3.711} )[/tex]
[tex]P(X < 20 ) = P( \frac{X - \mu }{ \sigma } < 0.4904 )[/tex]
Generally
[tex]\frac{X - \mu }{ \sigma } = Z ( The \ standardized \ value \ of \ X )[/tex]
So
[tex]P(X < 20 ) = P( Z< 0.4904 )[/tex]
From the z-table
[tex]P( Z< 0.4904 ) = 0.68807[/tex]
The probability is
[tex]P(X < 20 ) = 0.68807[/tex]
