Answer:
The probability is [tex]P(14 < X < 20 ) = 0.5354[/tex]
Step-by-step explanation:
From the question we are told that
The proportion that live with their parents is [tex]\r p = 0.142[/tex]
The sample size is n = 125
Given that there are two possible outcomes and that this outcomes are independent of each other then we can say the Recent census data follows a Binomial distribution
i.e
[tex]X \ \~ \ B( \mu , \sigma )[/tex]
Now the mean is evaluated as
[tex]\mu = n * \r p[/tex]
[tex]\mu = 125 * 0.142[/tex]
[tex]\mu = 17.75[/tex]
Generally the proportion that are not staying with parents is
[tex]\r q = 1 - \r p[/tex]
= > [tex]\r q = 0.858[/tex]
The standard deviation is mathematically evaluated as
[tex]\sigma = \sqrt{n * \r p * \r q }[/tex]
[tex]\sigma = \sqrt{ 125 * 0.142 * 0.858 }[/tex]
[tex]\sigma = 3.90[/tex]
Given the n is large then we can use normal approximation to evaluate the probability as follows
[tex]P(14 < X < 20 ) = P( \frac{ 14 - 17.75}{3.90} <\frac{ X - \mu }{\sigma } < \frac{ 20 - 17.75}{3.90} )[/tex]
Now applying continuity correction
[tex]P(14 < X < 20 ) = P( \frac{ 13.5 - 17.75}{3.90} < \frac{ X - \mu }{\sigma } < \frac{ 19.5 - 17.75}{3.90} )[/tex]
Generally
[tex]\frac{ X - \mu }{\sigma } = Z ( The \ standardized \ value \ of X )[/tex]
[tex]P(14 < X < 20 ) = P( \frac{ 13.5 - 17.75}{3.90} < Z< \frac{ 19.5 - 17.75}{3.90} )[/tex]
[tex]P(14 < X < 20 ) = P( -1.0897 < Z< 0.449 } )[/tex]
[tex]P(14 < X < 20 ) = P( Z< 0.449 ) - P(Z < -1.0897)[/tex]
So for the z - table
[tex]P( Z< 0.449 ) = 0.67328[/tex]
[tex]P(Z < -1.0897) = 0.13792[/tex]
[tex]P(14 < X < 20 ) = 0.67328 - 0.13792[/tex]
[tex]P(14 < X < 20 ) = 0.5354[/tex]
