Suppose a research firm conducted a survey to determine the mean amount steady smokers spend on cigarettes during a week. A sample of 100 steady smokers revealed that the sample mean is $20. The population standard deviation is $5. What is the probability that a sample of 100 steady smokers spend between $19 and $21?
A)0.4772
B)0.0228
C)0.9544
D)1.0000

Question

Quizplus · Accepted Answer

First, we need to standardize the sample mean using the formula z = (x̄ - μ) / (σ / √n):

z(lower) = (19 - 20) / (5 / √100) = -2

z(upper) = (21 - 20) / (5 / √100) = 2

Next, we can use a standard normal distribution table to find the probability that a z-score falls between -2 and 2. This is equivalent to finding the area under the standard normal curve between z = -2 and z = 2.

Using the table or a calculator, we can find that the probability is 0.9544.

Therefore, the answer is C.

Suppose a Research Firm Conducted a Survey to Determine the Mean