Question

 

Suppose that miles driven anually by cars in America are normally distributed with mean = 12; 894 miles and standard deviation = 1190 miles.

(a)If one car is chosen at random, what is the probability it has driven more than

13,000 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 13,000 miles?

***A parameter is a value for a population, and a statistic

is a value for a sample.

T F

Answer 1

 

Given that ,

mean = = 12894

standard deviation = = 1190

a) P( x > 13000) = 1 -P(x < 13000 )

= 1 - P[(x - ) / < (13000 - 12894) /1190 ]

=1 - P(z < 0.09)

= 1 - 0.5359 = 0.4641

Probability = 0.4641

b)

n = 25

= = 12894

= / n = 1190 / 25 = 238

P( < 13000 ) = P(( - ) / < (13000 - 12894) /238 )

= P(z < 0.45)

= 0.6736

Probability = 0.6736

F