Question

The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 44,663 miles, with a standard deviation of 2594 miles. What is the probability that the sample mean would differ from the population mean by less than 199 miles in a sample of 209 tires if the manager is correct?

Answer 1

Answer:

The probability is  $P(| \= X- \mu| < 199 ) = 0.733$

Step-by-step explanation:

From the question we are told that

   The mean mileage of a tire is   $\mu = 44663$

   The standard deviation is  $\sigma = 2594$

   The sample size is  n  =  209

 Generally the standard error of mean is mathematically represented as

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

=>  $\sigma_{x} = \frac{2594}{\sqrt{209} }$  

=>  $\sigma_{x} = 179.4$      

Generally  the probability that the sample mean would differ from the population mean by less than 199 miles  is mathematically represented as

     $P(| \= X- \mu| < 199 ) = P(|\frac{\= X - \mu}{ \sigma_{x}}| < \frac{199}{ 179.4})$

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

=>  $P(| \= X- \mu| < 199 ) = P(|Z| < \frac{199}{ 179.4})$

=>   $P(| \= X- \mu| < 199 ) = P(|Z|< 1.11)$

=>   $P(| \= X- \mu| < 199 ) = P(-1.11 \le Z \le 1.11 )$

=>   $P(| \= X- \mu| < 199 ) = P(Z \le 1.11 ) - P( Z \le -1.11 )$

From the z  table  the area under the normal curve to the left corresponding to 1.11 and -1.11 is  

        $P(Z \le 1.11 ) = 0.8665$

and

       $P(Z \le - 1.11 ) = 0.1335$

So

    $P(| \= X- \mu| < 199 ) = 0.8665 - 0.1335$

=>  $P(| \= X- \mu| < 199 ) = 0.733$