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
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 find the probability that the baker made enough cookies (i.e., that 70 cookies will be sufficient to meet demand), we will use the properties of the normal distribution.
Given the following information:
- Average sales (mean) \((\mu)\) = 92 cookies
- Standard deviation \((\sigma)\) = 9 cookies
- Number of cookies made = 70 cookies
We need to find the probability that sales are less than or equal to 70 cookies:
1. First, we need to standardize the value of 70 cookies using the Z-score formula:
\[
Z = \frac{X - \mu}{\sigma}
\]
where:
- \(X\) = 70 (the number of cookies made),
- \(\mu\) = 92,
- \(\sigma\) = 9.
Calculating the Z-score:
\[
Z = \frac{70 - 92}{9} = \frac{-22}{9} \approx -2.44
\]
2. Next, we'll use the Z-score to find the corresponding probability. We can look up the Z-score in the standard normal distribution table or use a calculator for normal distributions.
For a Z-score of -2.44, we look up the cumulative probability (or use a calculator):
The cumulative probability \(P(Z < -2.44) \approx 0.0074\).
3. Finally, we need to convert this into a percentage and round it to the nearest tenth of a percent:
\[
P(Z < -2.44) \approx 0.0074 \times 100 \approx 0.74\%
\]
Thus, the probability that 70 cookies were enough for that day’s sales is approximately **0.7%** when rounded to the nearest tenth of a percent.
Given the following information:
- Average sales (mean) \((\mu)\) = 92 cookies
- Standard deviation \((\sigma)\) = 9 cookies
- Number of cookies made = 70 cookies
We need to find the probability that sales are less than or equal to 70 cookies:
1. First, we need to standardize the value of 70 cookies using the Z-score formula:
\[
Z = \frac{X - \mu}{\sigma}
\]
where:
- \(X\) = 70 (the number of cookies made),
- \(\mu\) = 92,
- \(\sigma\) = 9.
Calculating the Z-score:
\[
Z = \frac{70 - 92}{9} = \frac{-22}{9} \approx -2.44
\]
2. Next, we'll use the Z-score to find the corresponding probability. We can look up the Z-score in the standard normal distribution table or use a calculator for normal distributions.
For a Z-score of -2.44, we look up the cumulative probability (or use a calculator):
The cumulative probability \(P(Z < -2.44) \approx 0.0074\).
3. Finally, we need to convert this into a percentage and round it to the nearest tenth of a percent:
\[
P(Z < -2.44) \approx 0.0074 \times 100 \approx 0.74\%
\]
Thus, the probability that 70 cookies were enough for that day’s sales is approximately **0.7%** when rounded to the nearest tenth of a percent.