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?
Step 1. Calculate the z-score for $45.00.



Step 2. Give the probability for step 1. %



Step 3. Calculate the z-score for $65.00.



Step 4. Give the probability for step 3. %



Step 5. Add the probabilities from steps 1 & 3.

Answer 1

Answer:

0.6826

Step-by-step explanation:

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

Sample size n = 54

Sample mean will follow normal with mean = 55 and std error = $\frac{10}{\sqrt{54} } \\=2.205$

When x =45, $z=\frac{45-55}{10} =-1$

Step 1: z score = -1

Step 2:  P(-1<Z<0) = $0.3413$

Step 3 : Z score for 65 = $\frac{65-55}{10}=1$

Step 4: P(0<Z<1) = 0.3413

step 5:

P(45<x<55) = 0.6826