# 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

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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