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

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

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)  =  (\sigma)/(√(n) )`

=>
`\sigma_(x)  =  (2594)/(√(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(|(\= X - \mu)/( \sigma_(x))| < (199)/( 179.4))`

`(\= X -\mu)/(\sigma )  =  Z (The  \ standardized \  value\  of  \ \= X )`

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