Question

Suppose X is a normally distributed random variable with mean 0 and variance 1 (i.e., standard normal).

Compute the following (Hint: you can use the R functions pnorm and qnorm to answer these questions).
P r(X < 1:96)
P r(X > 1:64)
P r(0:5 < X < 0:5) 1% quantile, q:01 and 99% quantile, q:99 ? 5% quantile, q:05 and 95% quantile,

Answer 1

`P(X <1.96) = 0.975`

`P(X >1.64) = 0.0505`

`P(0.5 < X < 0.5) = 0`

Explanation:

Given

`\bar x = 0` --- Mean

`\sigma^2 = 1` --- Variance

Calculate the standard deviation

`\sigma^2 = 1`

`\sigma = 1`

Solving (a): P(X < 1.96)

First, we calculate the z score using:

`z = (X - \bar x)/(\sigma)`

This gives:

`z = (1.96 - 0)/(1)`

`z = (1.96)/(1)`

`z = 1.96`

The probability is then solved using:

`(X < 1.96) = P(z <1.96)`

From the standard normal distribution table

`P(z <1.96) = 0.97500`

So:

`P(X <1.96) = 0.975`

Solving (b): P(X > 1.64)

First, we calculate the z score using:

`z = (X - \bar x)/(\sigma)`

This gives:

`z = (1.64 - 0)/(1)`

`z = (1.64)/(1)`

`z = 1.64`

The probability is then solved using:

`(X > 1.64) = P(z >1.64)`

`P(z >1.64) = 1 - P(z<1.64)`

From the standard normal distribution table

`P(z >1.64) = 1 - 0.9495`

`P(z >1.64) = 0.0505`

So:

`P(X >1.64) = 0.0505`

Solving (c): P(0.5 < X < 0.5)

This can be split as:

`P(0.5 < X < 0.5) = P(0.5<X) - P(X<0.5)`

In probability:

`P(0.5<X) = 1 - P(X>0.5)`

`P(0.5<X) = 1 - [1 - P(X<0.5)]`

`P(0.5<X) = 1 - 1 + P(X<0.5)`

`P(0.5<X) = P(X<0.5)`

`P(0.5 < X < 0.5) = P(0.5<X) - P(X<0.5)` becomes

`P(0.5 < X < 0.5) = P(X<0.5) - P(X<0.5)`

`P(0.5 < X < 0.5) = 0`