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Suppose a research firm conducted a survey to determine the mean amount steady smokers spend on cigarettes during a week. A sample of 170 steady smokers revealed that the sample mean is $20. The population standard deviation is $5. What is the probability that a sample of 170 steady smokers spend between $19 and $21

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Answer:

0.9910 = 99.10% probability that a sample of 170 steady smokers spend between $19 and $21

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 z-score of a measure X is given by:


Z = (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 p-value, 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 = (\sigma)/(√(n)).

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

Mean of 20, standard deviation of 5:

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

Sample of 170:

This means that
n = 170, s = (5)/(√(170))

What is the probability that a sample of 170 steady smokers spend between $19 and $21?

This is the p-value of Z when X = 21 subtracted by the p-value of Z when X = 19.

X = 21


Z = (X - \mu)/(\sigma)

By the Central Limit Theorem


Z = (X - \mu)/(s)


Z = (21 - 20)/((5)/(√(170)))


Z = 2.61


Z = 2.61 has a p-value of 0.9955

X = 19


Z = (X - \mu)/(s)


Z = (19 - 20)/((5)/(√(170)))


Z = -2.61


Z = -2.61 has a p-value of 0.0045

0.9955 - 0.0045 = 0.9910

0.9910 = 99.10% probability that a sample of 170 steady smokers spend between $19 and $21

User Milan Solanki
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