Question

Assume that the weekly household expenditure in the statistical area of Croydon South, Victoria is normally distributed with a mean of $1,773 and a standard deviation of $461. What is the probability that a household's weekly expenditure is less than $2,225?

Answer 1

To find the probability that a household's weekly expenditure is less than $2,225 in Croydon South, Victoria, we can use the z-score formula and standard normal distribution. The probability is approximately 0.834, or 83.4%.

Step-by-step explanation:

To find the probability that a household's weekly expenditure is less than $2,225, we can use the z-score formula and standard normal distribution. First, we need to calculate the z-score using the formula:

z = (x - μ) / σ

Where x is the value we want to find the probability for, μ is the mean, and σ is the standard deviation. Substituting the given values, we have:

z = (2225 - 1773) / 461 = 0.974

Next, we can use a z-table or calculator to find the probability associated with the z-score. Looking up a z-score of 0.974, we find that the probability is approximately 0.834. Therefore, the probability that a household's weekly expenditure is less than $2,225 is 0.834, or 83.4%.