The nicotine content in a single cigarette of a particular brand has a distribution with mean 0.3 mg and standard deviation 0.1 mg. If 100 randomly selected cigarettes of this brand are analyzed, what is the probability that the resulting sample mean nicotine content will be less than 0.29

15.87% probability that the resulting sample mean nicotine content will be less than 0.29

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 the resulting sample mean nicotine content will be less than 0.29

This is the pvalue of Z when X = 0.29. So

By the Central Limit Theorem

has a pvalue of 0.1587

15.87% probability that the resulting sample mean nicotine content will be less than 0.29

dont think so

step-by-step explanation:

.632

step-by-step explanation:

6 * 1/10 = 3/5

3*0.01 = 0.03

3/5 + 0.03 = .63

2*0.002 = .002

.63 + 0.002 = .632