The weights of ice cream cartons are normally distributed with a mean weight of 7 ounces and a standard deviation of 0.4 ounce.
​(a) What is the probability that a randomly selected carton has a weight greater than 7.12 ​ounces?
​(b) A sample of 36 cartons is randomly selected. What is the probability that their mean weight is greater than 7.12 ​ounces?
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Part 1
​(a) The probability is 
  
enter your response here.
​(Round to four decimal places as​ needed.)

We are dealing with a normal distribution where the weights of the ice cream cartons follow \( N(7, 0.4) \) with a mean \( \mu = 7 \) ounces and a standard deviation \( \sigma = 0.4 \) ounces.

### (a) Probability of a randomly selected carton having a weight greater than 7.12 ounces

First, we need to calculate the **z-score** for 7.12 ounces. The z-score formula is:

\[
z = \frac{x - \mu}{\sigma}
\]

where:
- \( x = 7.12 \) (the weight we are interested in),
- \( \mu = 7 \) (mean), 
- \( \sigma = 0.4 \) (standard deviation).

Substitute the values:

\[
z = \frac{7.12 - 7}{0.4} = \frac{0.12}{0.4} = 0.3
\]

Now, we use a z-table (or a calculator) to find the probability that a carton weighs less than 7.12 ounces, corresponding to \( P(Z < 0.3) \).

Looking up \( z = 0.3 \) in the z-table, we get approximately:

\[
P(Z < 0.3) \approx 0.6179
\]

To find the probability that a carton weighs **more** than 7.12 ounces, we subtract this value from 1:

\[
P(Z > 0.3) = 1 - 0.6179 = 0.3821
\]

Thus, the probability that a randomly selected carton weighs more than 7.12 ounces is:

\[
\boxed{0.3821}
\]

### (b) Probability that the mean weight of 36 cartons is greater than 7.12 ounces

Now, we deal with a **sample mean** of 36 cartons. When sampling, the standard deviation of the sample mean (called the standard error) is calculated as:

\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}
\]

where:
- \( \sigma = 0.4 \) (population standard deviation),
- \( n = 36 \) (sample size).

Substitute the values:

\[
\sigma_{\bar{x}} = \frac{0.4}{\sqrt{36}} = \frac{0.4}{6} = 0.0667
\]

Now, we compute the z-score for the sample mean using the new standard error:

\[
z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} = \frac{7.12 - 7}{0.0667} = \frac{0.12}{0.0667} \approx 1.8
\]

Using the z-table for \( z = 1.8 \), we find:

\[
P(Z < 1.8) \approx 0.9641
\]

Thus, the probability that the sample mean is less than 7.12 ounces is approximately 0.9641. To find the probability that the sample mean is **greater** than 7.12 ounces, we subtract this value from 1:

\[
P(Z > 1.8) = 1 - 0.9641 = 0.0359
\]

Thus, the probability that the mean weight of 36 cartons is greater than 7.12 ounces is:

\[
\boxed{0.0359}
\]

### Summary:
- (a) The probability that a randomly selected carton has a weight greater than 7.12 ounces is **0.3821**.
- (b) The probability that the mean weight of 36 cartons is greater than 7.12 ounces is **0.0359**.

Let me know if you'd like further details or explanations! Here are some related questions for further exploration:

1. What is the probability that a carton weighs less than 6.9 ounces?
2. How does increasing the sample size affect the probability in part (b)?
3. What is the z-score for a carton weighing exactly 6.5 ounces?
4. How would the probabilities change if the standard deviation were 0.5 ounces?
5. What is the probability that the mean weight of a sample of 50 cartons is less than 6.9 ounces?

**Tip:** When working with sample means, the larger the sample size, the smaller the standard error, leading to a more precise estimate of the population mean.

The weights of ice cream cartons are normally distributed with a mean weight of 7 ounces and a standard deviation of 0.4 ounce. (a) What is the probability that a randomly selected carton has a weight greater than 7.12 ounces? (b) A sample of 36 cartons is randomly selected. What is the probability that their mean weight is greater than 7.12 ounces?

Find the probability that a randomly selected ice cream carton weighs more than 7.12 ounces, and the probability that the mean weight of 36 cartons exceeds 7.12 ounces. This problem involves key statistical concepts like normal distribution, z-scores, and sampling distribution.

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