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:
\[ \boxed{68.08} \]
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