Question

the weight of ice cream cartons are normally distributed with a mean weight of 13 ounces and a standard deviation of 0.6 ounce.

a) what is the probability that a randomly selected carton has a weight greater than 13.22 ounces?

b) a sample of 25 cartons are randomly selected. what is the probability that their mean weight is greater than 13.22 ounces?

Answer 1

µ = 13, σ = 0.6, n = 25

a) P(X > 13.22) =

= P( (X-µ)/σ > (13.22-13)/0.6)

= P(z > 0.3667)

= 1 - P(z < 0.3667)

Using excel function:

= 1 - NORM.S.DIST(0.3667, 1)

= 0.3569

b) P(X̅ > 13.22) =

= P( (X̅-μ)/(σ/√n) > (13.22-13)/(0.6/√25) )

= P(z > 1.8333)

= 1 - P(z < 1.8333)

Using excel function:

= 1 - NORM.S.DIST(1.8333, 1)

= 0.0334

Answer 2

To solve these probability questions, we will use the properties of the normal distribution and the Central Limit Theorem.

Given information: Mean weight (μ) of ice cream cartons = 13 ounces Standard deviation (σ) of ice cream cartons = 0.6 ounces

a) Probability that a randomly selected carton has a weight greater than 13.22 ounces:

We need to find the area under the normal curve to the right of 13.22 ounces. To do this, we will use the standard normal distribution (Z-distribution) by converting the value 13.22 to a Z-score.

Z-score = (X - μ) / σ

where X is the value we are interested in (13.22 ounces), μ is the mean (13 ounces), and σ is the standard deviation (0.6 ounces).

Z-score = (13.22 - 13) / 0.6 ≈ 0.37

Now, we find the probability associated with the Z-score using a standard normal distribution table or calculator. The probability of a Z-score of 0.37 is approximately 0.6443.

So, the probability that a randomly selected carton has a weight greater than 13.22 ounces is approximately 0.6443 or 64.43%.

b) Probability that the mean weight of a sample of 25 cartons is greater than 13.22 ounces:

For this part, we will use the Central Limit Theorem, which states that the sample means of sufficiently large samples from any population will be approximately normally distributed.

The mean of the sample means (μx̄) is the same as the population mean (μ), which is 13 ounces.

The standard deviation of the sample means (σx̄), also known as the standard error of the mean, is calculated using the formula:

σx̄ = σ / √n

where σ is the population standard deviation (0.6 ounces) and n is the sample size (25 cartons).

σx̄ = 0.6 / √25 = 0.6 / 5 = 0.12

Now, we need to find the probability that the mean weight of a sample of 25 cartons is greater than 13.22 ounces. We use the Z-distribution again by converting the value 13.22 to a Z-score.

Z-score = (X - μ) / σx̄

Z-score = (13.22 - 13) / 0.12 ≈ 1.83

Now, find the probability associated with the Z-score using a standard normal distribution table or calculator. The probability of a Z-score of 1.83 is approximately 0.9664.

So, the probability that the mean weight of a sample of 25 cartons is greater than 13.22 ounces is approximately 0.9664 or 96.64%.