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mario62 [17]
3 years ago
14

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?
Mathematics
1 answer:
kicyunya [14]3 years ago
6 0

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  

     

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