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

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

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) = (\sigma )/(√(n) )`

=>
`\sigma_(x) = (6)/(√(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((X - \mu )/(\sigma_(x)) > (103.7  -  106)/(1.044 )  )`

Generally
`(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`