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A survey found that women spend, on average, $146.21 on beauty products during the summer months, with a standard deviation of $29.44. What percentage of women spend less than $160 ? Round to the nearest hundredth. Assume the distribution is normal.

A survey found that women spend, on average, $146.21 \$ 146.21 on beauty products during the summer months, with a standard deviation of $29.44 \$ 29.44 . What percentage of women spend less than $160 \$ 160 ? Round to the nearest hundredth. Assume the distribution is normal.

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Q. A survey found that women spend, on average, $146.21 \$ 146.21 on beauty products during the summer months, with a standard deviation of $29.44 \$ 29.44 . What percentage of women spend less than $160 \$ 160 ? Round to the nearest hundredth. Assume the distribution is normal.
  1. Identify Mean and Standard Deviation: Identify the mean and standard deviation of the spending on beauty products. The mean (average) spending is given as \$\(146\).\(21\), and the standard deviation is \$\(29\).\(44\).
  2. Calculate Z-Score for \(\$160\): Calculate the z-score for the spending amount of \(\$160\). The z-score is calculated using the formula: \(z = \frac{X - \mu}{\sigma}\), where \(X\) is the value we are comparing to the mean (\(\mu\)), and \(\sigma\) is the standard deviation. For \(X = \$160\), \(\mu = \$146.21\), and \(\sigma = \$29.44\), the z-score is: \(z = \frac{(\$160 - \$146.21)}{\$29.44}\) \(\$160\)\(0\)
  3. Find Percentile in Normal Distribution: Use the z-score to find the corresponding percentile in the standard normal distribution. We look up the z-score of \(0.468\) in a standard normal distribution table or use a calculator that provides the cumulative distribution function (CDF) for the normal distribution. The percentile for \(z = 0.468\) is approximately \(0.6808\) or \(68.08\%\).
  4. Interpret Percentile: Interpret the percentile.\(\newline\)Since we are looking for the percentage of women who spend less than \(\$160\), and the z-score corresponds to the area to the left of \(\$160\) in the normal distribution, the percentile we found is the answer to our question.

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