Answer:
3.67% probability of randomly seleting 37 cigarettes with a mean of 0.809 g or less.
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:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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 question, we have that:
![\mu = 0.895, \sigma = 0.292, n = 37, s = \frac{0.292}{\sqrt{37}} = 0.048](https://tex.z-dn.net/?f=%5Cmu%20%3D%200.895%2C%20%5Csigma%20%3D%200.292%2C%20n%20%3D%2037%2C%20s%20%3D%20%5Cfrac%7B0.292%7D%7B%5Csqrt%7B37%7D%7D%20%3D%200.048)
Find the probability of randomly seleting 37 cigarettes with a mean of 0.809 g or less.
This is the pvalue of Z when X = 0.809.
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
By the Central Limit Theorem
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{0.809 - 0.895}{0.048}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B0.809%20-%200.895%7D%7B0.048%7D)
![Z = -1.79](https://tex.z-dn.net/?f=Z%20%3D%20-1.79)
has a pvalue of 0.0367
3.67% probability of randomly seleting 37 cigarettes with a mean of 0.809 g or less.