Question

2. The mean birth weight of a full term boy born in the US is 7.7 lbs, with standard deviation of 1.1 lbs. A. Find the probability that a boy born full term in the US weighs less than 7 lbs. B. Find the probability that the average weight of 25 baby boys born in a particular hospital have an average weight less than 7 lbs.

Answer 1

Answer:

A

The value is $P( X < 7 ) =0.26226$

B

The value is $P( X < 7 ) = 0.00073$

Step-by-step explanation:

From the question we are told that

  The population mean is  $\mu = 7.7 lbs$

   The standard deviation is $\sigma = 1.1 \ lbs$

  The sample size is  n =  25

Generally the probability that a boy born full term in the US weighs less than 7 lbs is mathematically represented as

     $P( X < 7 ) = P( \frac{X - \mu }{\sigma} < \frac{7 - 7.7 }{1.1 } )$

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

=>   $P( X < 7 ) = P(Z <- 0.6364 )$

From the z table  the area under the normal curve to the left corresponding to    -0.6364  is

    $P(Z <- 0.6364 ) =0.26226$

=>   $P( X < 7 ) =0.26226$

Generally the standard error of mean is mathematically represented as  

       $\sigma_{x} = \frac{\sigma }{\sqrt{n} }$

=>     $\sigma_{x} = \frac{ 1.1 }{\sqrt{25} }$

=>     $\sigma_{x} = 0.22$

Generally the probability that the average weight of 25 baby boys born in a particular hospital have an average weight less than 7 lbs is mathematically represented as

         $P( \= X < 7 ) = P( \frac{\= X - \mu }{\sigma_{x}} < \frac{7 - 7.7 }{ 0.22 } )$

=>      $P( X < 7 ) = P(Z <- 3.1818 )$

From the z table  the area under the normal curve to the left corresponding to    -3.1818  is

=>      $P(Z <- 3.1818 ) = 0.00073$

=>      $P( X < 7 ) = 0.00073$