Question
A bakery owner wants to ensure they make enough cookies each day to meet the demand from customers. On average, they sell 12 cookies a day with a standard deviation of 9. The baker makes 70 coolues each day. Using a calculater or a spreadsheet program. And 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
Answers
GPT-4o mini
To find the probability that the bakery sells no more than 70 cookies a day, we first need to standardize the number using the z-score formula:
\[
z = \frac{(X - \mu)}{\sigma}
\]
Where:
- \(X\) = the value we're interested in (70 cookies)
- \(\mu\) = the average number of cookies sold per day (12 cookies)
- \(\sigma\) = the standard deviation (9 cookies)
Plugging in the values:
\[
z = \frac{(70 - 12)}{9} = \frac{58}{9} \approx 6.44
\]
The z-score of 6.44 is quite high, indicating that 70 cookies is significantly above the average. We can use a z-table, a calculator, or a software tool to assess the probability of a z-score being less than 6.44.
For a z-score of 6.44, the corresponding probability (P(Z < 6.44)) is effectively 1 (or 100%), because such a z-score is so far into the tail of the normal distribution.
Thus, the probability of the bakery selling 70 cookies or fewer is:
\[
P(X \leq 70) \approx 100\%
\]
Rounded to the nearest tenth of a percent, the final answer is:
\[
\text{Probability of selling no more than 70 cookies is } 100.0\%
\]
\[
z = \frac{(X - \mu)}{\sigma}
\]
Where:
- \(X\) = the value we're interested in (70 cookies)
- \(\mu\) = the average number of cookies sold per day (12 cookies)
- \(\sigma\) = the standard deviation (9 cookies)
Plugging in the values:
\[
z = \frac{(70 - 12)}{9} = \frac{58}{9} \approx 6.44
\]
The z-score of 6.44 is quite high, indicating that 70 cookies is significantly above the average. We can use a z-table, a calculator, or a software tool to assess the probability of a z-score being less than 6.44.
For a z-score of 6.44, the corresponding probability (P(Z < 6.44)) is effectively 1 (or 100%), because such a z-score is so far into the tail of the normal distribution.
Thus, the probability of the bakery selling 70 cookies or fewer is:
\[
P(X \leq 70) \approx 100\%
\]
Rounded to the nearest tenth of a percent, the final answer is:
\[
\text{Probability of selling no more than 70 cookies is } 100.0\%
\]