Question

Suppose that the weight of seedless watermelons is normally distributed with mean 7 kg. and standard deviation 1.3 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 7.8 kg?  

d. What is the probability that a randomly selected watermelon will weigh more than 7.3 kg?  

e. What is the probability that a randomly selected seedless watermelon will weigh between 6.3 and 7.3 kg?  

f. The 90th percentile for the weight of seedless watermelons is ? kg

Answer 1

Let X be the weight of a randomly selected seedless watermelon.

We have given that weight of seedless watermelons is normally distributed with mean 7 kg and standard deviation 1.3 kg.

1. The distribution of X ~ N(7, 1.3)
2. If distribution is Normal then mean and median are same. Therefore the median of seedless watermelon weight is 7 kg.
3. Z-score for a seedless watermelon weighing 7.8 kg = (x – μ)/σ = (7.8 – 7)/1.3 = 0.6154

Z score = 0.6154

4. Now we want to find the probability that a randomly selected watermelon will weigh more than 7.3 kg.

We will solve this problem by standardising.

P( X > 7.3) = (X - µ)/ σ > (7.3 - 7)/1.3)

                              = P(Z > 0.231)

                              = 1 - P(Z ≤ 0.23)

                              = 1 – 0.5910 …… (Using statistical table)

                              = 0.409

The probability that a randomly selected watermelon will weigh more than 7.3 kg is 0.409.

5. Now we want to find the probability that a randomly selected seedless watermelon will weigh between 6.3 and 7.3 kg.

P(6.3 < X < 7.3) = P((6.3 – 7)/1.3 < (X - µ)/ σ <(7.3 - 7)/1.3)

                              = P(-0.538 < Z < 0.231)

                              = P(Z ≤ 0.23) – P(Z ≤ -0.54)

                              = 0.5910 – 0.2946 …… (Using statistical table)

                              = 0.2964

P(6.3 <X < 7.3) = 0.2964

The probability that a randomly selected seedless watermelon will weigh between 6.3 and 7.3 kg is 0.2964.

6. P₉₀ is the 90th percentile means the 90% of values less than C.

We want to find the value of C where 90% of the population values lies below it.

P(X<C) = 0.90

P((X – μ)/σ < (C – μ)/σ) =0.90

P(Z < (C – μ)/σ)) = 0.90

We use the standard normal distribution table to find the area under the curve closet to 0.90 and we can determine the Z score for 0.90.

The area 0.90 corresponding to the Z score is 1.28

P(Z < (C – μ)/σ)) = 0.90

ɸ ((C – µ)/σ) = ɸ (1.28)

(C - µ)/σ = 1.28

C = 1.28 * σ + µ

C = 1.28 * 1.3 + 7.

C = 8.664

C = 8.664

There are 90% scores less than 8.664

P₉₀ = 8.664

P₉₀ = 8.664

The 90th percentile for the weight of seedless watermelons is 8.664 kg.