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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,

Respuesta :

MrRoyal

Answer:

[tex]P(X <1.96) = 0.975[/tex]

[tex]P(X >1.64) = 0.0505[/tex]

[tex]P(0.5 < X < 0.5) = 0[/tex]

Step-by-step explanation:

Given

[tex]\bar x = 0[/tex] --- Mean

[tex]\sigma^2 = 1[/tex] --- Variance

Calculate the standard deviation

[tex]\sigma^2 = 1[/tex]

[tex]\sigma = 1[/tex]

Solving (a): P(X < 1.96)

First, we calculate the z score using:

[tex]z = \frac{X - \bar x}{\sigma}[/tex]

This gives:

[tex]z = \frac{1.96 - 0}{1}[/tex]

[tex]z = \frac{1.96}{1}[/tex]

[tex]z = 1.96[/tex]

The probability is then solved using:

[tex](X < 1.96) = P(z <1.96)[/tex]

From the standard normal distribution table

[tex]P(z <1.96) = 0.97500[/tex]

So:

[tex]P(X <1.96) = 0.975[/tex]

Solving (b): P(X > 1.64)

First, we calculate the z score using:

[tex]z = \frac{X - \bar x}{\sigma}[/tex]

This gives:

[tex]z = \frac{1.64 - 0}{1}[/tex]

[tex]z = \frac{1.64}{1}[/tex]

[tex]z = 1.64[/tex]

The probability is then solved using:

[tex](X > 1.64) = P(z >1.64)[/tex]

[tex]P(z >1.64) = 1 - P(z<1.64)[/tex]

From the standard normal distribution table

[tex]P(z >1.64) = 1 - 0.9495[/tex]

[tex]P(z >1.64) = 0.0505[/tex]

So:

[tex]P(X >1.64) = 0.0505[/tex]

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

This can be split as:

[tex]P(0.5 < X < 0.5) = P(0.5<X) - P(X<0.5)[/tex]

In probability:

[tex]P(0.5<X) = 1 - P(X>0.5)[/tex]

[tex]P(0.5<X) = 1 - [1 - P(X<0.5)][/tex]

[tex]P(0.5<X) = 1 - 1 + P(X<0.5)[/tex]

[tex]P(0.5<X) = P(X<0.5)[/tex]

[tex]P(0.5 < X < 0.5) = P(0.5<X) - P(X<0.5)[/tex] becomes

[tex]P(0.5 < X < 0.5) = P(X<0.5) - P(X<0.5)[/tex]

[tex]P(0.5 < X < 0.5) = 0[/tex]

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