Answer :
Answer:
The probability is [tex]P(| \= X- \mu| < 199 ) = 0.733[/tex]
Step-by-step explanation:
From the question we are told that
The mean mileage of a tire is [tex]\mu = 44663[/tex]
The standard deviation is [tex]\sigma = 2594[/tex]
The sample size is n = 209
Generally the standard error of mean is mathematically represented as
[tex]\sigma_{x} = \frac{\sigma}{\sqrt{n} } [/tex]
=> [tex]\sigma_{x} = \frac{2594}{\sqrt{209} } [/tex]
=> [tex]\sigma_{x} = 179.4 [/tex]
Generally the probability that the sample mean would differ from the population mean by less than 199 miles is mathematically represented as
[tex]P(| \= X- \mu| < 199 ) = P(|\frac{\= X - \mu}{ \sigma_{x}}| < \frac{199}{ 179.4})[/tex]
[tex]\frac{\= X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ \= X )[/tex]
=> [tex]P(| \= X- \mu| < 199 ) = P(|Z| < \frac{199}{ 179.4})[/tex]
=> [tex]P(| \= X- \mu| < 199 ) = P(|Z|< 1.11)[/tex]
=> [tex]P(| \= X- \mu| < 199 ) = P(-1.11 \le Z \le 1.11 )[/tex]
=> [tex]P(| \= X- \mu| < 199 ) = P(Z \le 1.11 ) - P( Z \le -1.11 )[/tex]
From the z table the area under the normal curve to the left corresponding to 1.11 and -1.11 is
[tex]P(Z \le 1.11 ) = 0.8665[/tex]
and
[tex]P(Z \le - 1.11 ) = 0.1335[/tex]
So
[tex]P(| \= X- \mu| < 199 ) = 0.8665 - 0.1335[/tex]
=> [tex]P(| \= X- \mu| < 199 ) = 0.733[/tex]