Question

Suppose that X follows the normal distributionN(1,12).

1. What are the mathematical mean and variance ofX?

2. Suppose you observed the the following data ofX: 0.19, −0.50, 2.42, 1.66, 0.24, −1.39, 0.23, 1.95, 4.25, 2.18. Calculate the sample mean and variance and compare to the mathematical mean andvariance you find in part 1.

3. Suppose you have a new data point which is not observed. What is the probability that it is greater than the largest observation in part 2 or smaller than the smallest observation in part 2? You may use the CDF of a standard normal Φ(x) to express your final answer.

Answer 1

Solution:

1) Given that, X ~ N(1, 12)

The Normal distribution has the following standard form:

X ~ N(μ, σ^{​​​​​​2})

Where, μ is mean of X and σ^{​​​​​​2} is variance of X.

Therefore, Mean of X is 1 and variance of X is 12.

2) Sample mean is given by the formula,

Where, n is sample size, x_{​​​​​​i}​​​​​'s are sample values and x̄ is sample mean.

We have, n= 10

The sample mean is 1.123.

Sample variance is given as follows:

Where, n is sample size, x_{​​​​​​i}​​​​​'s are sample values and x̄ is sample mean.

The sample variance is 2.7719.

3) Largest observation in part 2 is 4.25 and smallest observation in part 2 is -1.39.

We have to obtain P(X > 4.25 or X < -1.39) .

P(X > 4.25 or X < -1.39) = P(X > 4.25 ⋃ X < -1.39)

Since, the events (X > 4.25) and (X < -1.39) are mutually exclusive events, therefore,

P(X > 4.25 ⋃ X < -1.39) = P(X > 4.25) + P(X < -1.39)

We know that if X ~ N(μ, σ^{​​​​​​2}) then,

We have, μ = 1 and σ = √12

Using "NORM.S.DIST" function of excel we get,

P(Z > 0.9382) = 0.1741 and P(Z < -0.6899) = 0.2451

Hence, probability that a new data point is greater than the largest observation in part 2 or smaller than the smallest observation in part 2 is 0.4192.

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