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

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Answer:

95.44% probability that a sample of 100 steady smokers spend between $19 and $21

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 normally distributed samples are solved using the z-score formula.

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

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

In this problem, we have that:

[tex]\mu = 20, \sigma = 5, n = 100, s = \frac{5}{\sqrt{100}} = 0.5[/tex]

What is the probability that a sample of 100 steady smokers spend between $19 and $21

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

X = 21

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{21 - 20}{0.5}[/tex]

[tex]Z = 2[/tex]

[tex]Z = 2[/tex] has a pvalue of 0.9772

X = 19

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{19 - 20}{0.5}[/tex]

[tex]Z = -2[/tex]

[tex]Z = -2[/tex] has a pvalue of 0.0228

0.9772 - 0.0228 = 0.9544

95.44% probability that a sample of 100 steady smokers spend between $19 and $21