Question

Alicia is a busy university student who spends her days in university attending classes or studying. In addition to her three meals, she packs two snacks to take with her to school. However, she doesn't always consume both snacks. Let X be the number of snacks she consumes per day. The distribution of X is as follows: x 0 1 2 P(X=x) 0.05 0.35 0.60 Assume the number of snacks per day is independent from day to day. A.[5] Find the sampling distribution of the average of snacks for two randomly selected days. B.[6] Find the expected value and variance for both X and �%. C.[3] Suppose that 36 days were selected at random. What is the sampling distribution of the sample mean based on n = 36? Why? D.[3] Suppose that 36 days were selected at random. What is the probability that the average number of snacks that Alicia consumed is at most 2.5 snacks per day, during the selected 36 days.

Answer 1

For two randomly selected days, the possible samples of X are,

X = 0, 0 with = 0 with p = 0.05 * 0.05 = 0.0025

X = 0, 1 and X = 1, 0 with = 0.5 with p = 0.05 * 0.35 +  0.35 * 0.05 = 0.035

X = 0, 2 and X = 2, 0 and X = 1,1 with = 1 with p = 0.05 * 0.60 + 0.60  * 0.05 + 0.35 * 0.35 = 0.1825

X = 1, 2 and X = 2, 1   with = 1.5 with p = 0.35 * 0.6 + 0.6 * 0.35 = 0.42

X = 2, 1 with = 2 with p = 0.6 * 0.6 = 0.36

The sampling distribution of the average of snacks for two randomly selected days is,

P( = 0) = 0.0025

P( = 0.5) = 0.035

P( = 1) = 0.1825

P( = 1.5) = 0.42

P( = 2) = 0.36

E(X) = 0.05 * 0 + 0.35 * 1 + 0.6 * 2 = 1.55

E(X²) = 0.05 * 0² + 0.35 * 1² + 0.6 * 2^{​​​​​​​2} = 2.75

Var(X) = E(X²) - E(X)² = 2.75 - 1.55² = 0.3475

E() = 0.0025 * 0 + 0.035 * 0.5 + 0.1825 * 1 + 0.42 * 1.5 + 0.36 * 2 = 1.55

E() = 0.0025 * 0² + 0.035 * 0.5² + 0.1825 * 1² + 0.42 * 1.5² + 0.36 * 2² = 2.57625

Var() = E() - E()²    = 2.57625 - 1.55² = 0.17375

Now, the sample size is greater than 30, so by Central limit theorem, the sampling distribution of the sample mean based on n = 36 follow Normal distribution, with mean as E(X) and variance as Var(X)/36

Mean = 1.55

Variance of sample mean = 0.3475/36

Standard deviation of sample mean = sqrt(0.3475/36 ) = 0.0982

Probability that the average number of snacks that Alicia consumed is at most 2.5 snacks per day = P( 2.5)

= P[Z (2.5 - 1.55)/0.0982]

= P[Z 9.67]

= 1