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A delivery driver has an average daily gasoline expense of $55.00. The standard deviation is $10.00. The owner takes a sample of 54 bills. What is the probability the mean of his sample will be between $45.00 and $65.00

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Answer:

Step-by-step explanation:

Given that a delivery driver has an average daily gasoline expense of $55.00. The standard deviation is $10.00.

As per central limit theorem, for samples of large size drawn at random, the sample mean follows a normal distribution irrespective of the original distribution.

Hence here we can say sample mean

[tex]\bar X =N(55,\frac{10}{\sqrt{54} } )[/tex]

= N(55,1.361)

required probability =

the probability the mean of his sample will be between $45.00 and $65.00

= [tex]P(45<\bar x<65)\\= F(65)-F(45)\\= 1[/tex]

i.e. almost a certain event.

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