Question

The American Cancer Society has stated that a 25-year-old man who smokes a pack of cigarettes a day gives up, on average, 5.5 years of life. Assuming that the number of years lost, x, is normally distributed with mean 5.5 and standard deviation 1.5, find:

a. the probability that the decrease in life of such a man is between 4 and 7 years.

b. the value of x such that 20% of the values lie above it.

Answer 1

Solution :

Given that ,

mean = = 5.5

standard deviation = = 1.5

(a)

P(4 < x < 7) = P((4 - 5.5)/ 1.5) < (x - ) / < (7 - 5.5) / 1.5) )

= P(-1 < z < 1)

= P(z < 1) - P(z < -1)

= 0.8413 - 0.1587 = 0.6826

Probability = 0.6826

(b)

P(Z > z) = 20%

1 - P(Z < z) = 0.20

P(Z < z) = 1 - 0.20 = 0.80

P(Z < 0.8416) = 0.80

z = 0.8416

Using z-score formula,

x = z * +

x = 0.8416 * 1.5 + 5.5 = 6.75

x = 6.76