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?

The probability that a randomly selected carton has a weight greater than 9.12 ounces is about 38%. The probability that the mean weight of 25 cartons is greater than 9.12 ounces is about 7%.

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Denote the random variable "weight of a randomly selected carton in ounces" by `X . ` It is given that `X ` is normally distributed with the mean `mu = 9 ` and the standard deviation `sigma = 0.4 .`

To answer the first part, convert `X ` to the standard normal random variable `Y = ( X - mu ) / sigma , ` then `X gt= 9.12 ` is equivalent to `Y gt= ( 9.12 - mu ) / sigma = 0.12 / 0.4 = 0.3 . ` Now we can use the normal distribution table to determine that `P ( Y gt= 0.3 ) approx 0.5 - 0.12 = 0.38. ` The number `0.12 ` is from the normal table for `0.3 , ` which means about `0.12 ` of values are between `0 ` and `0.3 . ` There are a half `(0.5) ` of non-negative values because the standard normal distribution is symmetric, so the part of interest has the probability of about `0.5 - 0.12.`

To answer the second part, recall that an averaged series of `n ` independent normal variables with the same parameters `mu ` and `sigma ` is again a normal random variable `bar X ` with the mean `mu ` and the standard deviation of `sigma / sqrt ( n ) . ` Here `n = 25, ` so the new standard deviation is `0.4 / 5 = 0.08 .`

Normalize `bar X ` to obtain `bar Y ` and the inequality `bar Y gt= 0.12 / 0.08 = 1.5 . ` Now the standard normal table gives the value of about `0.43, ` so the probability in question is about `0.5 - 0.43 = 0.07 .`

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