Question
A bakery owner wants to ensure they make enough cookies each day to meet the demand from customers. On average, they sell 92 cookies a day with a standard deviation of 9. The baker makes 70 cookies each day. Using a calculator or a spreadsheet program, find the probability that the baker made enough cookies (no more than 70 cookies sold that day). Round the answer to the nearest tenth of a percent.(1 point) The probability that 70 cookies were enough for that day’s sales is
Answers
GPT-4o mini
To determine the probability that the bakery will sell no more than 70 cookies, we can use the normal distribution. Given that the average (mean) number of cookies sold is 92 with a standard deviation of 9, we can standardize the value of 70 cookies using the Z-score formula:
\[
Z = \frac{X - \mu}{\sigma}
\]
where:
- \(X\) is the value we are interested in (70 cookies),
- \(\mu\) is the mean (92 cookies),
- \(\sigma\) is the standard deviation (9 cookies).
Substituting the values into the formula:
\[
Z = \frac{70 - 92}{9} = \frac{-22}{9} \approx -2.4444
\]
Next, we need to look up the Z-score of approximately -2.44 in the standard normal distribution table, or we can use a calculator or spreadsheet function to find the probability.
Using Excel, for example, the formula would be:
\[
= NORM.DIST(70, 92, 9, TRUE)
\]
Calculating this gives us the cumulative probability, which represents the probability of selling 70 cookies or fewer.
Using the standard normal distribution table or a calculator, we find:
\[
P(Z < -2.44) \approx 0.0074
\]
This means the probability of selling 70 cookies or fewer is approximately 0.0074.
To express this as a percentage, we multiply by 100:
\[
0.0074 \times 100 \approx 0.74\%
\]
Finally, rounding to the nearest tenth of a percent:
**The probability that 70 cookies were enough for that day’s sales is approximately 0.7%.**
\[
Z = \frac{X - \mu}{\sigma}
\]
where:
- \(X\) is the value we are interested in (70 cookies),
- \(\mu\) is the mean (92 cookies),
- \(\sigma\) is the standard deviation (9 cookies).
Substituting the values into the formula:
\[
Z = \frac{70 - 92}{9} = \frac{-22}{9} \approx -2.4444
\]
Next, we need to look up the Z-score of approximately -2.44 in the standard normal distribution table, or we can use a calculator or spreadsheet function to find the probability.
Using Excel, for example, the formula would be:
\[
= NORM.DIST(70, 92, 9, TRUE)
\]
Calculating this gives us the cumulative probability, which represents the probability of selling 70 cookies or fewer.
Using the standard normal distribution table or a calculator, we find:
\[
P(Z < -2.44) \approx 0.0074
\]
This means the probability of selling 70 cookies or fewer is approximately 0.0074.
To express this as a percentage, we multiply by 100:
\[
0.0074 \times 100 \approx 0.74\%
\]
Finally, rounding to the nearest tenth of a percent:
**The probability that 70 cookies were enough for that day’s sales is approximately 0.7%.**