Enter your answers as numbers accurate to 4 decimal places.

Question

A particular fruit's weights are normally distributed, with a mean of 536 grams and a standard deviation of 10 grams.

If you pick 24 fruits at random, what is the probability that their mean weight will be between 529 grams and 539 grams.

Enter your answers as numbers accurate to 4 decimal places.

Answer 1

We will use the sampling distribution of the sample mean.

μ = 536 grams (population mean)

σ = 10 grams (population standard deviation)

n = 24 fruits (sample size)

Calculate the standard error of the sample mean.

SE = σ / √n = 10 / √24 ≈ 2.0412

Find two different z-scores using the formula. Then use the table from the textbook or online to find the probability.

z₁ = (X₁ - μ) / SE = (529 - 536) / 2.0412 ≈ -3.43

z₂ = (X₂ - μ) / SE = (539 - 536) / 2.0412 ≈ 1.47

P(z₁ < -3.43) ≈ 0.0003

P(z₂ < 1.47) ≈ 0.9292

P(-3.43 < z < 1.47) ≈ 0.9292 - 0.0003 ≈ 0.9289

The probability that the mean weight of 24 randomly selected fruits will be between 529 and 539 grams is about 0.9289.