Question

The population of weights of a particular fruit is normally distributed, with a mean of 317 grams and a standard deviation of 29 grams. If 25 fruits are picked at random, then 5% of the time, their mean weight will be greater than how many grams? Round your answer to the nearest gram.

Answer 1

The mean of the sample distribution of sample means will be equal to the population mean, which is 317 grams. The standard deviation of the sample distribution (also called the standard error) will be equal to the population standard deviation divided by the square root of the sample size:

standard error = 29 / sqrt(25) = 5.8

To find the weight at which the mean weight of 25 fruits will be greater than 5% of the time, we need to find the z-score that corresponds to the 95th percentile (since 5% of the time we want the mean weight to be greater than some value). Using a standard normal distribution table or calculator, we find that the z-score corresponding to the 95th percentile is approximately 1.645.

Now we can use the formula for z-scores to find the weight at which the mean weight of 25 fruits will be greater than 5% of the time:

z = (x - mu) / se

where x is the weight we want to find, mu is the population mean (317 grams), and se is the standard error (5.8 grams).

Substituting the values we have:

1.645 = (x - 317) / 5.8

Solving for x, we get:

x = 317 + 1.645 * 5.8 = 326.5

Rounding to the nearest gram, we get that the mean weight of 25 fruits will be greater than 326 grams 5% of the time.