Question

The nicotine content in a single cigarette of a particular brand has a distribution with mean 0.9 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 0.89

Answer 1

Answer:

The answer to the question is;

The probability that the resulting sample mean of nicotine content will be less than 0.89 is 0.1587 or 15.87 %.

Step-by-step explanation:

The mean of the distribution = 0.9 mg

The standard deviation of the sample = 0.1 mg

The size of the sample = 100

The mean of he sample = 0.89

The z score for sample mean is given by

$Z =\frac{X-\mu}{\sigma/ \sqrt{n} }$ where

X = Mean of the sample

μ = Mean of the population

σ = Standard deviation of the population

Therefore Z = $\frac{0.89-0.90}{0.1/\sqrt{100} }$ = -1

From the standard probabilities table we have the probability for  a z value of -1.0 = 0.1587

Therefore the probability that the resulting sample mean will be less than 0.89 = 0.1587 That is the probability that the mean is will be less than 0.89 is 15.87 % probability.