Question

1) you are testing a new variety of watermelon developed at a research farm. You take a random sample of 16 watermelons from a field. From previous studies the population watermelon weight has been found to be 28.9 lb. The population standard deviation of weight was determined to be 5.6 lb. You can safely assume that the distribution of watermelon weights is close to normal.

a) what is the probability that your sample mean is more than 32 lb?

b) what is the probability that your sample mean is between 30 and 32 lb?

Answer 1

Let X be weight of watermelon.

X is normally distributed with mean (µ) 28.9 lb and Standard deviation σ = 5.6 lb

1. Now we want to find the probability that the sample mean is more than 32 lb.

We will solve this problem by standardising.

P( X > 32) = (X - µ)/ σ > (32 - 28.9)/5.6)

                              = P(Z > 0.55)

                              = 1 - P(Z ≤ 0.55)

                              = 1 - 0.7088 …… (Using statistical table)

                              = 0.2912

The probability that your sample mean is more than 32 lb is 0.2912.

2. Now we want to find the probability that the sample mean is between 30 lb and 32 lb.

P(30 < X < 32) = P((30 – 28.9)/5.6 < (X - µ)/ σ <(32- 28.9)/5.6)

                              = P(0.20 < Z < 0.55)

                              = P(Z ≤ 0.55) – P(Z ≤ 0.20)

                              = 0.7088 – 0.5793 …… (Using statistical table)

                              = 0.1295

P(30 <X < 32) = 0.1295

The probability that your sample mean is between 30 and 32 lb is 0.1295.