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
%.
To find the probability that the baker made enough cookies (no more than 70 sold), we can use the z-score formula and the standard normal distribution.
First, we calculate the z-score:
z = (70 - 92) / 9 = -2.44
Next, we look up the z-score in a standard normal distribution table or use a calculator to find the cumulative probability for a z-score of -2.44. This is approximately 0.0073.
Convert this probability to a percentage by multiplying by 100:
0.0073 * 100 = 0.73%
Therefore, the probability that the baker made enough cookies (no more than 70 cookies sold) is approximately 0.73%.