Question

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

Answer 1

Answer:

$P(X$

$P(X >1.64) = 0.0505$

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

Step-by-step 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:

$z = \frac{X - \bar x}{\sigma}$

This gives:

$z = \frac{1.96 - 0}{1}$

$z = \frac{1.96}{1}$

$z = 1.96$

The probability is then solved using:

$(X < 1.96) = P(z$

From the standard normal distribution table

$P(z$

So:

$P(X$

Solving (b): P(X > 1.64)

First, we calculate the z score using:

$z = \frac{X - \bar x}{\sigma}$

This gives:

$z = \frac{1.64 - 0}{1}$

$z = \frac{1.64}{1}$

$z = 1.64$

The probability is then solved using:

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

$P(z >1.64) = 1 - P(z$

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:

$P(0.5 < X < 0.5) = P(0.5$

In probability:

$P(0.50.5)$

$P(0.5$

$P(0.5$

$P(0.5$

$P(0.5 < X < 0.5) = P(0.5$ becomes

$P(0.5 < X < 0.5) = P(X$

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