Consider the the distribution of under-inflated tires on a four-wheel automobile. The mass

density is given by:

p(0) = 0.4, p(1) = p(2) = p(3) = 0.1, p(4) = 0.3

a) Find the expected value and the variance of this distribution.

b) For what proportion of such cars will the number of underinflated tires be within 1 standard deviation of the mean?

To find the expected value and variance of the distribution, we can use the formulas:

a) Expected Value:
The expected value is calculated by multiplying each possible value by its corresponding probability and summing them up.

E(X) = ΣXP(X)

Using the given mass density values, we have:
E(X) = (0 * 0.4) + (1 * 0.1) + (2 * 0.1) + (3 * 0.1) + (4 * 0.3)
E(X) = 0 + 0.1 + 0.2 + 0.3 + 1.2
E(X) = 1.8

Therefore, the expected value of the distribution is 1.8.

b) Variance:
The variance is calculated by subtracting the expected value from each possible value, squaring the result, multiplying by the corresponding probability, and summing them up.

Var(X) = Σ[P(X)*(X - E(X))^2]

Using the given mass density values and expected value, we have:
Var(X) = (0.4 * (0 - 1.8)^2) + (0.1 * (1 - 1.8)^2) + (0.1 * (2 - 1.8)^2) + (0.1 * (3 - 1.8)^2) + (0.3 * (4 - 1.8)^2)
Var(X) = (0.4 * (-1.8)^2) + (0.1 * (-0.8)^2) + (0.1 * (0.2)^2) + (0.1 * (1.2)^2) + (0.3 * (2.2)^2)
Var(X) = (0.4 * 3.24) + (0.1 * 0.64) + (0.1 * 0.04) + (0.1 * 1.44) + (0.3 * 4.84)
Var(X) = 1.296 + 0.064 + 0.004 + 0.144 + 1.452
Var(X) = 2.96

Therefore, the variance of the distribution is 2.96.

b) Proportion within 1 standard deviation of the mean:
To find the proportion of cars where the number of underinflated tires is within 1 standard deviation of the mean, we need to find the range of values within one standard deviation above and below the mean.

The standard deviation (σ) is the square root of the variance, which is √2.96 ≈ 1.72.

The range within one standard deviation of the mean is (mean - σ, mean + σ) = (1.8 - 1.72, 1.8 + 1.72) = (0.08, 3.52).

To find the proportion within this range, we need to sum up the probabilities of the values within this range:

P(X within range) = P(X = 1) + P(X = 2) + P(X = 3) = 0.1 + 0.1 + 0.1 = 0.3.

Therefore, for approximately 30% of cars, the number of underinflated tires will be within 1 standard deviation of the mean.