Question

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

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

Solution :

Given that,

mean = = 15.2

standard deviation = = 1.6

a ) Using standard normal table,

P(Z < z) = 3%

P(Z < z) = 0.03

P(Z < - 0.61) = 0.03

z = -1.88

Using z-score formula,

x = z * +

x = -1.88 * 1.6 + 15.2

x = 12.19

b ) P (x > 15.2 )

= 1 - P (x < 15.2 )

= 1 - P ( x -  / ) < ( 15.2 - 15.2 / 1.6)

= 1 - P ( z < 0 / 1.6 )

= 1 - P ( z < 0 )

Using z table

= 1 - 0.5000

= 0.5000

Probability = 0.5000

c ) n = 40

= 15.2

= / n = 1.6 40 = 0.2530

P (15.5 < x < 15.9 )

P ( 15.5 -15.2 / 0.2530) < ( x -  / ) < ( 15.9 - 15.2 / 0.2530)

P ( 0.3 / 0.2530 < z < 0.7 / 0.2530 )

P (1.18 < z < 2.77 )

P ( z < 2.77 ) - P ( z < 1.18 )

Using z table

=0.9972 - 0.8810

= 0.1162

Probability = 0.1162