A sample of n = 64 Tiger Swallowtail butterflies was collected independently fro

Question

A sample of n = 64 Tiger Swallowtail butterflies was collected independently from a population in Baja, California. It is known that the distribution of the wing length (X) has mean value E[X] = 4 cm and a variance of Var(X) = 20 square cm. (i)What is the probability that a randomly selected butterfly has a wing length between 3.5 and 4.5 cm? (ii)Compute the mean and variance of the sample mean (based on the n = 64 observation)? (iii) What is the (approximate) probability that the sample mean (based on the n = 64 observation) will fall between 3.5 and 4.5 cm? (iv) What is the (approximate) probability that the sample mean (based on the n = 64 observation) is smaller than 4.5 cm, given that we know that the sample mean is larger than 3.5? (v) Compute the mean and variance of the sample mean (based on the n = 225 observation)? (vi) What is the (approximate) probability that the sample mean (based on the n = 225 observation) will fall between 3.5 and 4.5 cm? (vii) What is the (approximate) probability that the sample mean (based on the n = 225 observation) is smaller than 4.5 cm. given that we know that the sample mean is larger than 3.5?

Explanation / Answer

Back-up Theory

If a random variable X ~ N(µ, 2), i.e., X has Normal Distribution with mean µ and variance 2, then

Z = (X - µ)/ ~ N(0, 1), i.e., Standard Normal Distribution ………………………..(1)

P(X or t) = P[{(X - µ)/ } or {(t - µ)/ }] = P[Z or {(t - µ)/ }] .………(2)

X bar ~ N(µ, 2/n),…………………………………………………………….…….(3),

where X bar is average of a sample of size n from population of X.

So, P(X bar or t) = P[Z or {(n)(t - µ)/ }] …………………………………(4)

Probability values for the Standard Normal Variable, Z, can be directly read off from

Standard Normal Tables or using Excel Function……………………………………..(5)

P(A/B) = P(AB)/P(B) ……………………………………………………………….(6)

Now, to work out solution,

Let X = wing length. We assume X ~ N(µ, 2).

We are given, µ = 4 and 2 = 20.

Part (i)

Probability a randomly selected butterfly has wing length between 3.5 and 4.5

= P(3.5 < X < 4.5).

[vide (2) under Back-up Theory] ,

P(3.5 < X < 4.5) = P[{(3.5 - 4)/25} < Z < {(4.5 - 4)/25}] = P(- 0.112 < Z < 0.112 )

P(Z < 0.112 ) - P(Z < - 0.112) = 0.5458 - 0.4542 = 0.0916 ANSWER

Part (ii) Given n = 64

[vide (3) under Back-up Theory] , sample mean Xbar ~ N(4, 20/64)

=> Mean of the sample mean is 4 and variance is 0.3125 ANSWER

Part (iii)

Probability the sample mean will fall between 3.5 and 4.5 = P(3.5 < Xbar < 4.5).

[vide (4) under Back-up Theory] ,

P(3.5 < Xbar < 4.5) = P[{(3.5 - 4)/0.559} < Z < {(4.5 - 4)/0.559}] = P(- 0.894 < Z < 0.894 )

P(Z < 0.894 ) - P(Z < - 0.894) = 0.8144 - 0.1856 = 0.6288 ANSWER

Part (iv)

Probability the sample mean is smaller than 4.5 given that it is larger than 3.5

= P(Xbar < 4.5/Xbar > 3.5) = P(3.5 < Xbar < 4.5)/P(Xbar > 3.5) [vide (6) under Back-up Theory] ,

= 0.6288/0.9144 = 0.6877 ANSWER

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