Question

Treatment providers for problem gamblers report that men who approached them for intervention had lost a mean of $2,849 in the preceding four weeks, according to a 2002 report. Assume that the distribution of gambling losses is normally distributed with a mean of $2,849 and a standard deviation of $900. We would expect to see 95% of the number of gambling losses to be between what two values? A.1049 and 4649 B. 1949 and 3749 C. 149 and 5549

Answer 1

To find the range of values within which we would expect to see 95% of the number of gambling losses, we can calculate the z-score and use the formula Range = mean ± (z-score * standard deviation). The range is approximately $1,054.4 to $4,643.6.

Step-by-step explanation:

To find the range of values within which we would expect to see 95% of the number of gambling losses, we can use the concept of z-scores and the standard normal distribution. In this case, we have a mean of $2,849 and a standard deviation of $900. Since we want to find the range that contains 95% of the values, we need to find the z-score that corresponds to the cumulative probability of 0.025 (from -0.025 to 0.025 on both sides of the mean). Using a z-table or a calculator, we can find that the z-score is approximately 1.96.

The formula to find the range is:

Range = mean ± (z-score * standard deviation)

Plugging in the values, we get:

Range = $2,849 ± (1.96 * $900)

Calculating the values, we get:

Lower value = $2,849 - (1.96 * $900) ≈ $1,054.4

Upper value = $2,849 + (1.96 * $900) ≈ $4,643.6

Therefore, we would expect to see 95% of the number of gambling losses to be between approximately $1,054.4 and $4,643.6. The correct answer is A. 1049 and 4649.