Question

For a normal distribution, boxes of "Craizy Raizy Bran" have a μ = 750 raisins, with a σ = 15. Find the probability that an individual box has less than 720 raisins.

Answer 1

To find the probability that an individual box has less than 720 raisins in a normal distribution with a mean (μ) of 750 raisins and a standard deviation (σ) of 15, we can use the z-score formula to standardize the value of 720. The probability is approximately 0.0228, or 2.28%.

Step-by-step explanation:

To find the probability that an individual box has less than 720 raisins in a normal distribution with a mean (μ) of 750 raisins and a standard deviation (σ) of 15, we need to standardize the value of 720 using the z-score formula.

The z-score formula is: z = (x - μ) / σ

Plugging in the values, we get: z = (720 - 750) / 15 = -2

We can then use a standard normal distribution table or a calculator with a normal distribution function to find the probability corresponding to a z-score of -2.

The probability is approximately 0.0228, or 2.28%.