The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.972 g and a standard deviation of 0.289 g. The company that produces these cigarettes claims that it has now reduced the amount of nicotine. The supporting evidence consists of a sample of 36 cigarettes with a mean nicotine amount of 0.929 g. 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) =

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:

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

In this problem, we have that:

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

By the Central Limit Theorem

has a pvalue of 0.1867. So

P( ¯ x < 0.929 g) = 0.1867

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)