Question

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

Answer 1

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

$\bar X =N(55,\frac{10}{\sqrt{54} } )$

= N(55,1.361)

required probability =

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

= $P(45$

i.e. almost a certain event.