A particular fruit's weights are normally distributed...

Question

A particular fruit's weights are normally distributed, with a mean of 224 grams and a standard deviation of 16 grams. If you pick 11 fruit at random, what is the probability that their mean weight will be between 209 grams and 214 grams?

Answer 1

The distribution of the mean weight of 11 fruit picked at random will also be normally distributed, with a mean of 224 grams and a standard deviation of 16 grams divided by the square root of 11 (the square root of the sample size).

Let's define the random variable X as the mean weight of a sample of 11 fruit. Then,

X ~ N(224, 16/sqrt(11))

We want to find the probability that X is between 209 and 214 grams. We can standardize X using the z-score formula:

z = (X - μ) / (σ / sqrt(n))

where μ is the population mean (224 grams), σ is the population standard deviation (16 grams), and n is the sample size (11).

So,

z = (209 - 224) / (16/sqrt(11)) = -2.62 z = (214 - 224) / (16/sqrt(11)) = -1.46

Using a standard normal table or calculator, we can find that the probability of z being between -2.62 and -1.46 is about 0.0446.

Therefore, the probability that the mean weight of 11 fruit picked at random will be between 209 grams and 214 grams is approximately 0.0446 or 4.46%.