Question

A particular brand of cigarettes has a mean nicotine content of 15.2 mg with standard deviation of 1.6mg.

a)What percent of cigarettes has a mean nicotine content less than 15mg?
b) What is the probability that a random sample of 20 of these cigarettes has a mean nicotine content of more than 16 mg?
c)What is a mean nicotine content for lowest 3% of all cigarettes?
d) What is the probability randomly chosen cigarette has a nicotine content greater than 15.2mg?
e) What is the probability that a random sample of 40 of these cigarettes has a mean nicotine content between 15.5 and 15.9 mg?

Answer 1

Mean, = 15.2 mg

Standard deviation, = 1.6 mg

(a) The percent of cigarettes has a mean nicotine content less than 15 mg = P(X < 15)

= P{Z < (15 - 15.2)/1.6}

= P(Z < -0.125)

= 0.4502 = 45.02%

(b) Corresponding to n = 20, the standard error = 1.6/√20 = 0.358

The probability that the mean is greater than 16 mg

= P{Z > (16 - 15.2)/0.358}

= P(Z > 2.236)

= 0.0127

(c) Corresponding to lowest 3% of nicotine, the z value = -1.881

The required mean nicotine content = 15.2 - 1.881*1.6 = 12.19 mg

(d) The required probability = P(X > 15.2) = P(Z > 0) = 0.5

(e) For n = 40, the standard error = 1.6/√40 = 0.253

The required probability = P(15.5 < < 15.9)

= P{(15.5 - 15.2)/0.253 < Z < (15.9 - 15.2)/0.253}

= P(1.186 < Z < 2.767)

= 0.115