Question

a)

Suppose that miles driven anually by cars in America are normally distributed with mean μ =
12, 394 miles and standard deviation σ = 1190 miles.
(a)If one car is chosen at random, what is the probability it has driven less than
12,000 miles last year?

(b) If a sample of 25 cars is taken, what is the probability that the mean of the
sample is less than 12,000 miles?

c)For a standard normal distribution, find P(1.22 < z < 2.51).

Answer 1

Solution :

Given that ,

mean = = 12394

standard deviation = = 1190

(a)

P(x < 12000) = P[(x - ) / < (12000 - 12394) / 1190]

= P(z < -0.33)

= 0.3707

(b)

= / n = 1190 / 25 = 238

P( < 12000) = P(( - ) / < (12000 - 12394) / 238)

= P(z < -1.66)

= 0.0485

(c)

P(1.22 < z < 2.51) = P(z < 2.51) - P(z < 1.22) = 0.994 - 0.8888 = 0.1052