Chapter 6: Q 33. (page 373)

Exercises 31–33 refer to the following setting. Choose an American household at random and
let the random variable X be the number of cars (including SUVs and light trucks) they own.
Here is the probability distribution if we ignore the few households that own more than 5cars.
About what percentage of households have a number of cars within 2 standard deviations
of the mean?
a. 68%
b. 71%
c. 93%
d. 95%
e. 98%

Short Answer

Expert verified

$93 %$ percentage of households have a number of cars within 2 standard deviations

of the mean.

Step by step solution

Step 1. Given information

We have given probability distribution:

Step 2. To find the standard deviation.

The expected value is the sum of the product of the each possibility with its probability.

$μ = \sum_{} x P (x) = 0 \times 0.09 + 1 \times 0.36 + 2 \times 0.35 + 3 \times 0.13 + 4 \times 0.05 + 5 \times 0.02 = 1.75$

The variance is the expected value of square deviation.

$σ^{2} = \sum_{} {(x - μ)}^{2} P (x) = {(0 - 1.75)}^{2} \times 0.09 + {(1 - 1.75)}^{2} \times 0.36 + {(2 - 1.75)}^{2} \times 0.35 + {(3 - 1.75)}^{2} \times 0.13 + {(4 - 1.75)}^{2} \times 0.05 + {(5 - 1.75)}^{2} \times 0.02 = 1.1675$

The standard deviation is $σ = \sqrt{σ^{2}} = \sqrt{1.1675} \approx 1.0805$

Step 3. To find 2 standard deviations of the mean.

$μ - 2 σ = 1.75 - 2 (1.0805) = - 0.4110 μ + 2 σ = 1.75 + 2 (1.0805) = 3.9110$

Now,

$P (μ - 2 σ < X < μ + 2 σ) = P (- 0.4110 < X < 3.9110) P (0 \leq X \leq 3) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) = 0.09 + 0.36 + 0.35 + 0.13 = 0.93 = 93 %$