Question

Recent census data indicated that 14.2% of adults between the ages of 25 and 34 live with their parents. A random sample of 125 young adults in this age group was selected. What is the probability that between 14 and 20 of these young adults lived with their parents?

Answer 1

Answer:

The  probability is  $P(14 < X < 20 ) = 0.5354$  

Step-by-step explanation:

From the question we are told that

   The  proportion that live with their parents is  $\r p = 0.142$

   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  

       $X \ \~ \ B( \mu , \sigma )$

Now the mean is evaluated as

      $\mu = n * \r p$

      $\mu = 125 * 0.142$

      $\mu = 17.75$

Generally the proportion that are not staying with parents is  

      $\r q = 1 - \r p$

= >    $\r q = 0.858$

The standard deviation is mathematically evaluated as

     $\sigma = \sqrt{n * \r p * \r q }$

     $\sigma = \sqrt{ 125 * 0.142 * 0.858 }$

    $\sigma = 3.90$

Given the n is large  then we can use normal approximation to evaluate the probability as follows  

     $P(14 < X < 20 ) = P( \frac{ 14 - 17.75}{3.90}$

Now applying continuity correction

      $P(14 < X < 20 ) = P( \frac{ 13.5 - 17.75}{3.90} < \frac{ X - \mu }{\sigma } < \frac{ 19.5 - 17.75}{3.90} )$

Generally  

    $\frac{ X - \mu }{\sigma } = Z ( The \ standardized \ value \ of X )$

    $P(14 < X < 20 ) = P( \frac{ 13.5 - 17.75}{3.90} < Z< \frac{ 19.5 - 17.75}{3.90} )$

     $P(14 < X < 20 ) = P( -1.0897 < Z< 0.449 } )$

    $P(14 < X < 20 ) = P( Z< 0.449 ) - P(Z < -1.0897)$

So  for the z -  table  

         $P( Z< 0.449 ) = 0.67328$

         $P(Z < -1.0897) = 0.13792$

 $P(14 < X < 20 ) = 0.67328 - 0.13792$    

  $P(14 < X < 20 ) = 0.5354$