Final Answer:
The probability that a sample of 100 steady smokers spend between $76 and $84 is 95.44%.
Step-by-step explanation:
To find the probability that a sample of 100 steady smokers spends between $76 and $84 on cigarettes during a week, we would use the normal distribution as the sampling distribution of the sample mean can be approximated by a normal distribution due to the Central Limit Theorem (since the sample size is large, n=100).
We are given:
The sample mean x is $80.
The population standard deviation σ is $20.
The sample size (n) is 100.
We're looking for P(76 < X < 84), where X is the sample mean of the distribution of sample means.
First, we need to find the standard error of the mean, which will tell us the standard deviation of the sampling distribution of the sample mean.
The standard error (SE) is calculated by dividing the population standard deviation by the square root of the sample size:
SE = σ/{√n}
= {20}/{\√{100}}
= {20}/{10} = 2
Next, we want to convert our sample mean range into z-scores. A z-score is a measure of how many standard deviations an element is from the mean.
The formula to calculate a z-score is:
z = {(X - x)}/{SE}
For the lower bound ($76):
z_{76} = {(76 - 80)}/{2} = {-4}{2} = -2
For the upper bound ($84):
z_{84} = {(84 - 80)}/{2} = 2
Now we check a z-table (or use a calculator that can compute probabilities for the normal distribution) to find out the probabilities corresponding to z-scores of -2 and 2.
The probability for z ≤ -2 is approximately 0.0228, and the probability for z ≤ 2 is approximately 0.9772.
The probability of being between z = -2 and z = 2 is the probability at the upper bound minus the probability at the lower bound:
P(-2 < Z < 2) = P(Z < 2) - P(Z < -2)
P(-2 < Z < 2) = 0.9772 - 0.0228
P(-2 < Z < 2) = 0.9544
So, the probability that a sample of 100 steady smokers spends between $76 and $84 on cigarettes during a week is approximately 95.44%.