Determine P(-k≤X≤k) when X follows the standard normal distribution - High school Statistics & Probabilities Exercises

X follows \mathcal{N}\left(0{,}1\right).

Determine P\left(-1.4 \lt X \lt 1.4\right) using the following table.

P\left(-1.4 \lt X \lt 1.4\right)=0.7\ 576

P\left(-1.4 \lt X \lt 1.4\right)=0.808

P\left(-1.4 \lt X \lt 1.4\right)=0.9\ 192

P\left(-1.4 \lt X \lt 1.4\right)=0.8\ 384

For a normal distribution with mean \mu=0 and standard deviation \sigma=1, we have:

P\left(-a \lt X \lt a\right)=P\left(X \lt a\right)-P\left(X \lt -a\right)

And:

P\left(X \lt -a\right)=1-P\left(X \lt a\right)

Therefore:

P\left(-a \lt X \lt a\right)=P\left(X \lt a\right)-P\left(X \lt -a\right)=P\left(X \lt a\right)-\left(1-P\left(X \lt a\right)\right)=2\cdot P\left(X \lt a\right)-1

According to the table:

P\left(X \lt 1.4\right)=0.9\ 192

Thus:

P\left(-1.4 \lt X \lt 1.4\right)=2\cdot P\left(X \lt 1.4\right)-1

P\left(-1.4 \lt X \lt 1.4\right)=2\cdot0.9\ 192-1

P\left(-1.4 \lt X \lt 1.4\right)=0.8\ 384

X follows \mathcal{N}\left(0{,}1\right).

Determine P\left(-2.3 \lt X \lt 2.3\right) using the following table.

Insert a table of a normal distribution.

P\left(-2.3 \lt X \lt 2.3\right)=0.9\ 281

P\left(-2.3 \lt X \lt 2.3\right)=0.9\ 218

P\left(-2.3 \lt X \lt 2.3\right)=0.9\ 786

P\left(-2.3 \lt X \lt 2.3\right)=0.9\ 182

For a normal distribution with mean \mu=0 and standard deviation \sigma=1, the z score of any value in the distribution equals the value z_{a}=a and we have:

P\left(-a \lt X \lt a\right)=P\left(X \lt a\right)-P\left(X \lt -a\right)

Since:

P\left(X \lt -a\right)=1-P\left(X \lt a\right)

We have:

P\left(-a \lt X \lt a\right)=P\left(X \lt a\right)-P\left(X \lt -a\right)=P\left(X \lt a\right)-\left(1-P\left(X \lt a\right)\right)=2\cdot P\left(X \lt a\right)-1

In our problem:

P\left(X \lt 2.3\right)=0.9\ 893

So:

P\left(-2.35 \lt X \lt 2.3\right)=2\cdot P\left(X \lt 2.3\right)-1=2\cdot0.9\ 893 -1 = 0.9\ 786

P\left(-2.35 \lt X \lt 2.35\right)=0.9\ 812

X follows \mathcal{N}\left(0{,}1\right).

Determine P\left(-1.76 \lt X \lt 1.76\right) using the following table.

Insert a table of a normal distribution.

P\left(-1.76 \lt X \lt 1.76\right)=0.9\ 612

P\left(-1.76 \lt X \lt 1.76\right)=0.9\ 621

P\left(-1.76 \lt X \lt 1.76\right)=0.8\ 575

P\left(-1.76 \lt X \lt 1.76\right)=0.9\ 216

For a normal distribution with mean \mu=0 and standard deviation \sigma=1, the z score of any value in the distribution equals the value:

z_{a}=a

Also:

P\left(-a \lt X \lt a\right)=P\left(X \lt a\right)-P\left(X \lt -a\right)

Since:

P\left(X \lt -a\right)=1-P\left(X \lt a\right)

We have:

P\left(-a \lt X \lt a\right)=P\left(X \lt a\right)-P\left(X \lt -a\right)=P\left(X \lt a\right)-\left(1-P\left(X \lt a\right)\right)=2\cdot P\left(X \lt a\right)-1

In our problem:

P\left(X \lt 1.76\right)=0.9\ 608

Thus:

P\left(-1.76 \lt X \lt 1.76\right)=2\cdot P\left(X \lt 1.76\right)-1=2\cdot0.9\ 608-1=0.9\ 216

P\left(-1.76 \lt X \lt 1.76\right)=0.9\ 216

X follows \mathcal{N}\left(0{,}1\right).

Determine P\left(-0.5 \lt X \lt 0.5\right) using the following table.

Insert a table of a normal distribution.

P\left(-0.5 \lt X \lt 0.5\right)=0.4\ 129

P\left(-0.5 \lt X \lt 0.5\right)=0.398

P\left(-0.5 \lt X \lt 0.5\right)=0.1\ 417

P\left(-0.5 \lt X \lt 0.5\right)=0.2\ 174

For a normal distribution with mean \mu=0 and standard deviation \sigma=1, the z score of any value in the distribution equals the value:

z_{a}=a

Also:

P\left(-a \lt X \lt a\right)=P\left(X \lt a\right)-P\left(X \lt -a\right)

Since:

P\left(X \lt -a\right)=1-P\left(X \lt a\right)

We have:

P\left(-a \lt X \lt a\right)=P\left(X \lt a\right)-P\left(X \lt -a\right)=P\left(X \lt a\right)-\left(1-P\left(X \lt a\right)\right)=2\cdot P\left(X \lt a\right)-1

In our problem:

P\left(X \lt 0.5\right)=0.5\ 199

Thus:

P\left(-0.5 \lt X \lt 0.5\right)=2\cdot P\left(X \lt 0.5\right)-1=2\cdot0.5\ 199-1=0.398

P\left(-0.5 \lt X \lt 0.5\right)=0.398

X follows \mathcal{N}\left(0{,}1\right).

Determine P\left(-2 \lt X \lt 2\right) using the following table.

Insert a table of a normal distribution.

P\left(-2 \lt X \lt 2\right)=0.2

P\left(-2 \lt X \lt 2\right)=0.8\ 572

P\left(-2 \lt X \lt 2\right)=0.8\ 147

P\left(-2 \lt X \lt 2\right)=0.9\ 544

For a normal distribution with mean \mu=0 and standard deviation \sigma=1, the z score of any value in the distribution equals the value:

z_{a}=a

Also:

P\left(-a \lt X \lt a\right)=P\left(X \lt a\right)-P\left(X \lt -a\right)

Since:

P\left(X \lt -a\right)=1-P\left(X \lt a\right)

We have:

P\left(-a \lt X \lt a\right)=P\left(X \lt a\right)-P\left(X \lt -a\right)=P\left(X \lt a\right)-\left(1-P\left(X \lt a\right)\right)=2\cdot P\left(X \lt a\right)-1

In our problem:

P\left(X \lt 2\right)=0.9\ 772

Thus:

P\left(-2 \lt X \lt 2\right)=2\cdot P\left(X \lt 2\right)-1=2\cdot0.9\ 772-1=0.9\ 544

P\left(-2 \lt X \lt 2\right)=0.9\ 544

X follows \mathcal{N}\left(0{,}1\right).

Determine P\left(-1.37 \lt X \lt 1.37\right) using the following table.

Insert a table of a normal distribution.

P\left(-1.37 \lt X \lt 1.37\right)=0.8\ 294

P\left(-1.37 \lt X \lt 1.37\right)=0.7\ 305

P\left(-1.37 \lt X \lt 1.37\right)=0.7\ 185

P\left(-1.37 \lt X \lt 1.37\right)=0.6\ 076

For a normal distribution with mean \mu=0 and standard deviation \sigma=1, the z score of any value in the distribution equals the value:

z_{a}=a

Also:

P\left(-a \lt X \lt a\right)=P\left(X \lt a\right)-P\left(X \lt -a\right)

Since:

P\left(X \lt -a\right)=1-P\left(X \lt a\right)

We have:

P\left(-a \lt X \lt a\right)=P\left(X \lt a\right)-P\left(X \lt -a\right)=P\left(X \lt a\right)-\left(1-P\left(X \lt a\right)\right)=2\cdot P\left(X \lt a\right)-1

In our problem:

P\left(X \lt 1.37\right)=0.9\ 147

Thus:

P\left(-1.37 \lt X \lt 1.37\right)=2\cdot P\left(X \lt 1.37\right)-1=2\cdot0.9\ 147-1=0.8\ 294

P\left(-1.37 \lt X \lt 1.37\right)=0.8\ 294