Question

A research firm conducted a study to determine the average amount of money that smokers spend on cigarettes during a week. The firm found that the population mean amount that all smokers spend on cigarettes is $20 and the population standard deviation is $5. What is the probability that a new sample this year of 100 steady smokers spends between $19 and $21 on average

Answer 1

Answer:

0.9544 = 95.44% probability that a new sample this year of 100 steady smokers spends between $19 and $21 on average

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The firm found that the population mean amount that all smokers spend on cigarettes is $20 and the population standard deviation is $5.

This means that $\mu = 20, \sigma = 5$

Sample of 100:

This means that $n = 100, s = \frac{5}{\sqrt{100}} = 0.5$

What is the probability that a new sample this year of 100 steady smokers spends between $19 and $21 on average?

This is the pvalue of Z when X = 21 subtracted by the pvalue of Z when X = 19. So

X = 21

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{21 - 20}{0.5}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772

X = 19

$Z = \frac{X - \mu}{s}$

$Z = \frac{19 - 20}{0.5}$

$Z = -2$

$Z = -2$ has a pvalue of 0.0228

0.9772 - 0.0228 = 0.9544

0.9544 = 95.44% probability that a new sample this year of 100 steady smokers spends between $19 and $21 on average