Question

According to PG&E Company, the amounts of utility bills for all households in a city have a skewed probability distribution with a mean of $140 and a standard deviation of $30. Find the probability that the mean amount of bills for a randomly selected sample of 75 households will be

1. within $6 of the population mean (hint: find two values of the sample mean)
2. more than the population mean by at least $4 (hint: find the value that the sample mean is greater than the population mean by at least $4)
3. The standard deviation of the mean amount of bills for a randomly selected sample of 75 households is

Answer 1

Given, μ = 140 and σ = 30

Sample size n = 75

Formula for standard deviation of the sampling distribution = σ/sqrt(n)

Thus, Std dev. of the mean amount of bills for a randomly selected sample of 75 households = 30/sqrt(75) = 3.4641

The probability that the mean amount of bills for a randomly selected sample of 75 households will be within $6 of the population mean = P(140-6 < μx̄ < 140+6) = P(134 < μx̄ < 146)

So, we need to find P(134 < μx̄ < 146)

P (-1.73 < Z < 1.73) = P( Z < 1.73) - P( Z < -1.73) = 0.9582 - 0.0418

P (-1.73 < Z < 1.73) = 0.9164

Thus, the probability that the mean amount of bills for a randomly selected sample of 75 households will be within $6 of the population mean is 0.9164

The probability that the mean amount of bills for a randomly selected sample of 75 households will be more than the population mean by at least $4 = P(μx̄ > 140+4) = P(μx̄ > 144)

So, we need to find P(μx̄ > 144)

Thus, the probability that the mean amount of bills for a randomly selected sample of 75 households will be more than the population mean by at least $4 is 0.1251