Question

The fuel efficiency in miles per gallon of all BMW 320i’s is approximately normally distributed with a mean of 25 and a standard deviation of 2. A dealer receives a shipment of a random sample of 320i’s (random with respect to mpg) from the factory. Hint: look at the sample sizes and think about which tables you’d need to use for these problems.

(a) Find the probability that average fuel efficiency is less than 24 mpg if the dealer receives one car.

(b) Repeat the exercises for four cars.

(c) Repeat for 16 cars and explain why your answers differ.

Answer 1

Given : mean ( µ ) = 25 and standard deviation (σ) = 2

According to sampling distribution of sample mean. The sample mean approximately follows normal distribution with mean $1585683651563_blob.png$ = $1585683651096_blob.png$ and standard deviation $1585683651089_blob.png$ = $1585683651115_blob.png$

a) n = 1 , so $1585683651563_blob.png$ = 25 and $1585683651089_blob.png$ = 2/√1 = 2

P( $\bar{x}$ < 24 ) = $P\left ( \frac{\bar{x}-\mu _{\bar{x}}}{\sigma _{\bar{x}}} \leq\frac{24-25}{2}\right )$ = P( z ≤ -0.5 )

P( $\bar{x}$ < 24 ) = 0.3085

b) n = 4 , so $1585683651563_blob.png$ = 25 and $1585683651089_blob.png$ = 2/√4 = 1

P( $\bar{x}$ < 24 ) = $P\left ( \frac{\bar{x}-\mu _{\bar{x}}}{\sigma _{\bar{x}}} \leq\frac{24-25}{1}\right )$ = P( z ≤ -1 )

P( $\bar{x}$ < 24 ) = 0.1587

c) n = 16 , so $1585683651563_blob.png$ = 25 and $1585683651089_blob.png$ = 2/√16 = 0.5

P( $\bar{x}$ < 24 ) = $P\left ( \frac{\bar{x}-\mu _{\bar{x}}}{\sigma _{\bar{x}}} \leq\frac{24-25}{2}\right )$ = P( z ≤ -2 )

P( $\bar{x}$ < 24 ) = 0.0228

As sample size n increases , standard deviation of $1585683651089_blob.png$ decreases , it leads to decrease in the probability.