Question

A survey by the National Retail Federation found that women spend on average $146.21 for the Christmas holidays. Assume the standard deviation is $29.44. Find the percentage of women who spend less than $160.00. Assume the variable is normally distributed. Express the answer in the form of a percentage (do not include % sign in your answer). Round your answer to the nearest hundredth.

Answer 1

To find the percentage of women who spend less than $160.00 during the Christmas holidays, we can use the properties of the normal distribution. Given that the average (mean) spending is \( \mu = 146.21 \) and the standard deviation is \( \sigma = 29.44 \), we can standardize the value of $160.00 to find its corresponding z-score.

The formula for calculating the z-score is:

\[
z = \frac{(X - \mu)}{\sigma}
\]

Where:
- \( X \) is the value we are interested in ($160.00),
- \( \mu \) is the mean ($146.21),
- \( \sigma \) is the standard deviation ($29.44).

Now, substituting the values:

\[
z = \frac{(160.00 - 146.21)}{29.44} \approx \frac{13.79}{29.44} \approx 0.468
\]

Next, we'll look up the z-score of 0.468 in the standard normal distribution table or use a calculator that provides the cumulative probability for z-scores.

The cumulative probability \( P(Z < 0.468) \) corresponds to the percentage of women who spend less than $160.00.

Using a standard normal distribution table or calculator, we find:

\[
P(Z < 0.468) \approx 0.6808
\]

To express this as a percentage, we multiply by 100:

\[
0.6808 \times 100 \approx 68.08
\]

Finally, rounding to the nearest hundredth, the percentage of women who spend less than $160.00 is:

\[
\boxed{68.08}
\]