Respuesta :
Answer:
0.0526 = 5.26% probability that 32 filtered cigarettes have a mean of 19.2 mg 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 normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
Given that the tar content of cigarettes have a mean of 20.1 mg and a standard deviation of 3.15 mg
This means that [tex]\mu = 20.1, \sigma = 3.15[/tex]
Sample of 32 filtered cigarettes
This means that [tex]n = 32, s = \frac{3.15}{\sqrt{32}} = 0.5568[/tex]
What is the probability that 32 filtered cigarettes have a mean of 19.2 mg or less?
This is the pvalue of Z when X = 19.2. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{19.2 - 20.1}{0.5568}[/tex]
[tex]Z = -1.62[/tex]
[tex]Z = -1.62[/tex] has a pvalue of 0.0526
0.0526 = 5.26% probability that 32 filtered cigarettes have a mean of 19.2 mg or less.
