Question

The nicotine content in a single cigarette of a particular brand has a distribution with mean 0.5 mg and standard deviation 0.1 mg. If 100 of these cigarettes are analyzed, what is the probability that the resulting sample mean nicotine content will be less than 0.49? (Round your answers to four decimal places.)

Answer 1

Answer:

0.1587

Explanation:

Given data

μ = 0.5 mg  

standard deviation σ =0.1 mg

n=100

we know that

$P(\overline X$

$Z= (\frac{\overline X-\mu}{\sigma/\sqrt{n} })$

putting the values we get

$Z= (\frac{0.49-0.5}{0.1/\sqrt{100} })$

Z=-1

Area under the curve for z =-1 is 0.1587 (from z score table)

P(X<0.49) = 0.1587

P(X<0.37) = 0.0013