Question

A racing car consumes a mean of 106 gallons of gas per race with a standard deviation of 6 gallons. If 33 racing cars are randomly selected, what is the probability that the sample mean would be greater than 103.7 gallons? Round your answer to four decimal places.

Answer 1

Answer:

The probability is  $P(\= X > 103.7 ) = 0.9862$

Step-by-step explanation:

From the question we are told that

  The population mean is  $\mu = 106 \ gallons$

   The population standard deviation is  $\sigma = 6 \ gallons$

    The sample size is  $n = 33$

   

Generally the standard deviation of sample mean is mathematically represented as  

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

=>      $\sigma_{x} = \frac{6}{\sqrt{33} }$

=>      $\sigma_{x} = 1.044$

Generally the probability that the sample mean would be greater than 103.7 gallons is mathematically represented as

        $P(\= X > 103.7 ) = P(\frac{X - \mu }{\sigma_{x}} > \frac{103.7 - 106}{1.044 } )$

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

So

$P(\= X > 103.7 ) = P(Z > -2.203 )$

From the z table  the probability of (Z > -2.203 ) is  

     $P(Z > -2.203 ) = 0.9862$

So

    $P(\= X > 103.7 ) = 0.9862$