Question

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.966 grams and a standard deviation of 0.315 grams. Find the probability of randomly selecting a cigarette with 0.305 grams of nicotine or less.

Answer 1

The probability is
`P(X \le 0.305 ) = 0.01795`

Explanation:

From the question we are told that

The population mean is
`\mu = 0.966 \ grams`

The standard deviation is
`\sigma = 0.315 \ grams`

Given that the amounts of nicotine in a certain brand of cigarette are normally distributed

Then the probability of randomly selecting a cigarette with 0.305 grams of nicotine or less is mathematically represented as

`P(X \le 0.305 ) = 1 - P(X > 0.305) = 1 - P((X - \mu )/(\sigma ) > (0.305 - \mu )/(\sigma ) )`

Generally

`(X - \mu )/(\sigma ) = Z (The \ standardized \ value \ of X )`

So

`P(X \le 0.305 ) = 1 - P(X > 0.305) = 1 - P(Z > (0.305 - 0.966 )/(0.315) )`

`P(X \le 0.305 ) = 1 - P(X > 0.305) = 1 - P(Z >-2.0984 )`

From the z-table(reference calculator dot net ) value of
`P(Z >-2.0984 ) =0.98205`

So

`P(X \le 0.305 ) = 1 - P(X > 0.305) = 1 - 0.98205`

=>
`P(X \le 0.305 ) = 1 - P(X > 0.305) = 0.01795`

=>
`P(X \le 0.305 ) = 0.01795`