Stats help?

Question

The nicotine content in a single cigarette of a particular brand has a distribution with mean 0.3 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 0.29?

Answer 1

Mean is μ =  0.30 mg .

Standarad deviation is σ = 0.1 mg .

Number samples taken under consideration n = 100

P(x < 0.29) = P (z < √n (x - μ)/σ  )

P(x < 0.29) = P (z < (√100)(0.29 - 0.30)/0.1 )

P(x < 0.29) = P (z < (10)(-0.01)/0.1 )

P(x < 0.29) = P (z < -0.1/0.1 ) ⇒ P (z < -1 )

P(x < 0.29) = P (z < -1 ) = 0.1587

[The values are taken from z-score table ]

Hence ,the probability that the resulting sample mean nicotine content will be less than 0.29 is 0.1587 .