# Suppose that the weight of see

Suppose that the weight of seedless watermelons is normallydistributed with mean 7 kg. and standard deviation 1.3 kg. Let X bethe weight of a randomly selected seedless watermelon. Round allanswers to 4 decimal places where possible.a. What is the distribution of X? X ~ N( , )b. What is the median seedless watermelonweight?  kg.c. What is the Z-score for a seedless watermelon weighing 7.8kg?  d. What is the probability that a randomly selected watermelon willweigh more than 7.3 kg?  e. What is the probability that a randomly selected seedlesswatermelon will weigh between 6.3 and 7.3 kg?  f. The 90th percentile for the weight of seedless watermelons is ?kg

Let X be the weight of a randomly selected seedlesswatermelon.

We have given that weight of seedless watermelons is normallydistributed 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 is7 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

1. Now we want to find the probability that a randomly selectedwatermelon will weigh more than 7.3 kg.

We will solve this problem bystandardising.

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 arandomly selected watermelon will weigh more than 7.3 kg is0.409.

1. Now we want to find the probability that a randomly selectedseedless 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 arandomly selected seedless watermelon will weigh between 6.3 and7.3 kg is 0.2964.

1. P90 is the 90th percentile means the 90% of valuesless than C.

We want to find the value of C where90% 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 normaldistribution table to find the area under the curve closet to 0.90and we can determine the Z score for 0.90.

The area 0.90 corresponding to the Zscore 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 than8.664

P90 = 8.664

P90 =8.664

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

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