A bakery owner wants to ensure they make enough cookies each day to meet the demand from customersOn 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 programfind the probability that the baker made enough cookies (no more than 70 cookies sold that day)Round the answer to the nearest tenth
To find the probability that the baker made enough cookies (no more than 70 sold), we need to calculate the z-score first.
z-score = (X - μ) / σ
where:
X = number of cookies sold (70)
μ = mean number of cookies sold (92)
σ = standard deviation (9)
z-score = (70 - 92) / 9
z-score = -22 / 9
z-score ≈ -2.44
Next, we look up the z-score in a standard normal distribution table or use a calculator to find the probability.
The probability of selling no more than 70 cookies is equal to the probability of having a z-score less than or equal to -2.44.
Using a standard normal distribution table or calculator, we find that the probability of z ≤ -2.44 is approximately 0.0073.
Therefore, the probability that the baker made enough cookies (no more than 70 sold) is approximately 0.0073, rounded to the nearest tenth.