Mathematics, 14.05.2021 22:00 edandjill24

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 from: Ehdhhxjdjjd866

0.9910 = 99.10% probability that a sample of 170 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 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:

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

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 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 $s = \frac{\sigma}{\sqrt{n}}$ .

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 $n = 170, s = \frac{5}{\sqrt{170}}$

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

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

By the Central Limit Theorem

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

$Z = \frac{21 - 20}{\frac{5}{\sqrt{170}}}$

Z = 2.61

Z = 2.61 has a p-value of 0.9955

X = 19

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

$Z = \frac{19 - 20}{\frac{5}{\sqrt{170}}}$

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

Answer from: Quest

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Answer from: Quest

is there supposed to be a diagram or a graph for this one? ?

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