Question

Suppose that the weight of seedless watermelons is normally distributed with mean 6.3 kg. and standard deviation 1.1 kg. Let X be the weight of a randomly selected seedless watermelon. Round all answers to 4 decimal places where possible.

a. What is the distribution of X? X ~ N( , )

b. What is the median seedless watermelon weight? kg.

c. What is the Z-score for a seedless watermelon weighing 6.9 kg?

d. What is the probability that a randomly selected watermelon will weigh more than 5.9 kg?

What is the probability that a randomly selected seedless watermelon will weigh between 5.7 and 6.6 kg?

The 70th percentile for the weight of seedless watermelons is kg.?

Answer 1

a) We are given the distribution here as:

b) For a normal distribution, median = mean that is 6.3 here.

Therefore median here is given as 6.3 kg

c) The z score here is computed as:

d) The required probability here is computed as:

P(X > 5.9)

Converting it to a standard normal variable, we get:

Getting it from the standard normal tables, we get:

Therefore 0.6419 is the required probability here.

e) The required probability here is computed as:

P( 5.7 < X < 6.6)

Converting it to a standard normal variable, we have:

Getting it from the standard normal tables, we get:

Therefore 0.3148 is the required probability here.

f) From standard normal tables, we have:

P(Z < 0.524) = 0.7

Therefore the 70th percentile value here is computed as:

= Mean + 0.524*Std Dev

= 6.3 + 0.524*1.1

= 6.8764

Therefore 6.8764 kg is the required 70^th percentile value required here.