The monthly expenditures on food by single adults living in one neighborhood of los angeles are normally distributed with a mean of $410 and a standard deviation of $75. determine the percentage of samples of size 9 that have mean expenditures within $20 of the population mean expenditure of $410.

0.5763

Step-by-step explanation:

We need to estimate the standard error of the mean.

Standard error of the mean = standard deviation of the original distribution/√sample size

Standard error of the mean = 75/√9 = 25

Now we can use this Standard error to estimate z as follows:

Z = (x – mean)/standard deviation

We want the mean expenditures within $20, so x – mean = 20 and -20

Z = (20)/25

Z = 0.8

Z = (-20)/25

Z = -0.8

Using  a Z table we can find probability  

P (-0.8<z<0.8)= 0.5763

step-by-step explanation:

if you want to solve percent error follow these steps.

1. subtract the accepted value from the experimental value.

2. take the absolute value of step 1.

3. divide that answer by the accepted value.

4. multiply that answer by 100 and add the % symbol to express the answer as a percentage.

if you want to solve percent of change just do this.

1. work out the difference (increase) between the two numbers you are comparing.

2. divide the increase by the original number and multiply the answer by 100. if your answer is a negative number then this is a percentage decrease.

hope this : )