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
[tex]Z =\frac{X-\mu}{\sigma/ \sqrt{n} }[/tex] where
X = Mean of the sample
μ = Mean of the population
σ = Standard deviation of the population
Therefore Z = [tex]\frac{0.89-0.90}{0.1/\sqrt{100} }[/tex] = -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.
