Suppose a research firm conducted a survey to determine the mean amount steady smokers spend on cigarettes during a week. Previous research indicates the population mean is $20 per week. The population standard deviation is $5. What is the probability that a sample of 100 steady smokers spend between $19 and $21

The probability is 0.9545

Step-by-step explanation:

To find this probability, we shall use the z-score formula

Mathematically;

z-score = (x - mean)/SD/√n

where mean = 20

SD = 5

n = 100 and so √100 = 10

So for 19, we have ;

(19-20)/5/10 = 1/0.5 = -2

For 21, we have

(21-20)/5/10 = 1/0.5 = 2

So the probability we want to calculate is

P( -2 < z < 2)

we can use the standard normal distribution table for this

Using this table;

P(-2<z<2) = 0.9545