A milk company unconditionally guarantees that an unopened, properly refrigerated container of its milk will be fresh for 5 days after the date stamped on the carton. The company's data shows that unopened containers remain fresh on average for 6 days, with a standard deviation of 0.65 days. Assume that the length of time milk stays fresh follows a normal distribution. What percent of the milk containers will spoil in greater than 7 days after the date stamped?

Question

A milk company unconditionally guarantees that an unopened, properly refrigerated container of its milk will be fresh for 5 days after the date stamped on the carton. The company's data shows that unopened containers remain fresh on average for 6 days, with a standard deviation of 0.65 days. Assume that the length of time milk stays fresh follows a normal distribution. What percent of the milk containers will spoil in greater than 7 days after the date stamped?

Answer 1

(7-6)/0.65 = 1.538 std

so, look that up in your Z table to see how much lies in the right tail.

Answer 2

To find the percentage of milk containers that will spoil in greater than 7 days after the date stamped, we need to calculate the z-score first.

Z-score formula: (x - μ) / σ

Where:
x = the value we want to find the percentage for (7)
μ = mean (6)
σ = standard deviation (0.65)

Calculating the z-score:

Z = (7 - 6) / 0.65
Z = 1.54 (rounded to two decimal places)

Now, we can use the z-score to find the corresponding percentage using a standard normal distribution table (or a calculator).

Looking up the z-score of 1.54 in the table, we find that the percentage is approximately 93.32%.

Therefore, approximately 93.32% of the milk containers will spoil in greater than 7 days after the date stamped.

Answer 3

To find the percentage of milk containers that will spoil in greater than 7 days after the date stamped, we need to calculate the area under the normal distribution curve to the right of 7.

To do this, we will use the z-score formula:

z = (x - μ) / σ

where:
- x is the value we want to find the percentage for (in this case, 7),
- μ is the mean of the distribution (6 days), and
- σ is the standard deviation of the distribution (0.65 days).

First, let's calculate the z-score:

z = (7 - 6) / 0.65 = 1.54

Now, we need to find the cumulative probability to the right of z = 1.54. Since the distribution is normal, we can use a standard normal distribution table or a calculator to find this probability.

Looking up the z-score of 1.54 in the standard normal distribution table, I found that the area to the left of z = 1.54 is approximately 0.9382. Therefore, the area to the right of z = 1.54 is 1 - 0.9382 = 0.0618.

So, approximately 0.0618 (or 6.18%) of the milk containers will spoil in greater than 7 days after the date stamped.