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

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 and standard deviation , the zscore of a measure X is given by:

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 and standard deviation , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation .

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:

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

By the Central Limit Theorem

has a pvalue of 0.9772

X = 19

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

4sqrt5

step-by-step explanation:

the base has dimensions of 8 x 4, so the diagonal is 64 + 16 square rooted or 4sqrt5.

312312

step-by-step explanation: