The weights of ice cream cartons are normally distributed with a mean weight of 9 ounces and a standard deviation of 0.4 ounce. What is the probability that a randomly selected carton has a weight greater than 9.12 ounces? A sample of 25 cartons is randomly selected. What is the probability that their mean weight is greater than 9.12 ounces?

a) 0.3821

b) 0.06681

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From the given information, we know that the weights of ice cream cartons X are normally distributed with a mean of 9 ounces and a standard deviation of 0.4 ounces. That is X~N(`\mu` = 9, `\sigma` = 0.4).

a) We want P(X > 9.12)

To find this probability, we need...

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From the given information, we know that the weights of ice cream cartons X are normally distributed with a mean of 9 ounces and a standard deviation of 0.4 ounces. That is X~N(`\mu` = 9, `\sigma` = 0.4).

a) We want P(X > 9.12)

To find this probability, we need to determine the Z score of X using the formula `Z = \frac{X - \mu}{\sigma}` .

Here, `Z = \frac{9.12 - 9}{0.4}` = 0.3

Using the standard normal distribution table

P(Z > 0.3) = 1 - P(Z < 0.3) = 1 - 0.61791 = 0.3821 (to 4 decimal places).

Alternatively, we can use the R statistical program to calculate this probability. This program is freely available at the following link: https://www.r-project.org/

The probability can be calculated by executing the following code in R:

> pnorm(9.12, mean=9, sd=0.4, lower.tail = FALSE)
[1] 0.3820886

b) We want P(`\bar{X}` > 9.12).

If X has a distribution with mean `\mu` and standard deviation `\sigma` and it is approximately normally distributed, then `\bar{X}` is approximately normally distributed with mean `\mu` and standard error `\frac{\sigma}{\sqrt{n}}` .

As such, `\bar{X}` ~ N(`\mu` , `\frac{\sigma}{\sqrt{n}}` ).

In this question, the sample size (n) = 25.

Thus, here, we find the Z score of `\bar{X}` as `\frac{9.12 - 9}{\frac{0.4}{\sqrt{25}}}` = `\frac{0.12}{0.08}` = 1.5.

Therefore, P(Z > 1.5) = 1 = P(Z < 1.5) = 1 - 0.93319 = 0.0668 (to 4 decimal places).

Alternatively, we can use the following R code to find this probability:

> pnorm(9.12, mean=9, sd=(0.4/5), lower.tail = FALSE)
[1] 0.0668072

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