Respuesta :
Answer:
The probability that the sample mean would differ from the population mean by greater than 2.4 gallons is 0.86558
Step-by-step explanation:
The given parameters are;
The mean, [tex]\bar x[/tex] = 101 gallons
The population mean = μ
The variance, σ² = 49
The number of racing cars, n = 43
The probability that the sample mean would differ from the population mean by greater than 2.4 gallons is given as follows;
P([tex]\left | \bar x - \mu \right |[/tex]) >2.4 = P(2.4 > [tex]\left | \bar x - \mu \right |[/tex] >-2.4)
The formula for the test of the mean, Z₀, is given as follows;
[tex]Z_0 = \dfrac{\bar x - \mu}{\dfrac{\sigma}{\sqrt{n} } }[/tex]
Therefore, we have;
[tex]P \left (\dfrac{2.4}{\dfrac{\sqrt{49} }{\sqrt{43} } } >Z> \dfrac{-2.4}{\dfrac{\sqrt{49} } {\sqrt{43} } } \right) = P \left (\dfrac{2.4}{\dfrac{7 }{\sqrt{43} } } >Z> \dfrac{-2.4}{\dfrac{7} {\sqrt{43} } } \right)[/tex]
Which gives;
P([tex]\left | \bar x - \mu \right |[/tex]) >2.4 = P(2.25 > Z > -2.25)
(We round up to 2.25 because that is the accuracy of the z-table)
From which we have;
P(Z > -2.25) - P(Z > 2.25) = (1 - P(Z < -2.25)) - (1 - P(Z < 2.25))
P([tex]\left | \bar x - \mu \right |[/tex]) >2.4 = (1 - 0.1222) - (1 - 0.98778) = 0.86558
The probability that the sample mean would differ from the population mean by greater than 2.4 gallons = 0.86558.
