Answer:
P( ¯ x < 0.929 g) = 0.1867
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 random variable X, with mean
and standard deviation
, the sample means with size n of at least 30 can be approximated to a normal distribution with mean
and standard deviation ![s = \frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
In this problem, we have that:
![\mu = 0.972, \sigma = 0.289, n = 36, s = \frac{0.289}{\sqrt{6}} = 0.0482](https://tex.z-dn.net/?f=%5Cmu%20%3D%200.972%2C%20%5Csigma%20%3D%200.289%2C%20n%20%3D%2036%2C%20s%20%3D%20%5Cfrac%7B0.289%7D%7B%5Csqrt%7B6%7D%7D%20%3D%200.0482)
Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly selecting 36 cigarettes with a mean of 0.929 g or less. P( ¯ x < 0.929 g) =
This is the pvalue of Z when X = 0.929. So
![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{0.929 - 0.972}{0.0482}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B0.929%20-%200.972%7D%7B0.0482%7D)
![Z = -0.89](https://tex.z-dn.net/?f=Z%20%3D%20-0.89)
has a pvalue of 0.1867. So
P( ¯ x < 0.929 g) = 0.1867