Determine P(a≤X≤b) when X follows any normal distribution and a and b are given Statistics & Probabilities

In order to find P\left(a \lt X \lt b\right) when X doesn't follow the standard normal distribution, we need to calculate the z -scores of a and b.

When \mu is the mean and \sigma is the standard deviation, the z -score is the number of standard deviations an observation lies from the mean:

• z_{a}=\dfrac{a-\mu}{\sigma}
• z_{b}=\dfrac{b-\mu}{\sigma}

In our problem, the mean is \mu=2 and the standard deviation is \sigma=3. We calculate the z -scores for -3 and 7 as:

• z_{-3}=\dfrac{-3-2}{3}\approx -1.67
• z_{7}=\dfrac{7-2}{3}\approx 1.67

The table of standard normal probabilities gives the probability of observing an outcome below the given z -score:

• P\left(X \lt a\right)\approx P\left(Z \lt z_{a}\right)
• P\left(X \lt b\right)\approx P\left(Z \lt z_{b}\right)

Looking at the standard normal probabilities table, we find P\left(Z \lt -1.67\right) and P\left(Z \lt 1.67\right).

• P\left(X \lt -3\right)\approx P\left(Z \lt -1.67\right)=1-P\left(Z \lt 1.67\right)\approx 1-0.9\ 525=0.475
• P\left(X \lt 7\right)\approx P\left(Z \lt 1.67\right)\approx 0.9\ 525

We know that:

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

Here we have:

P\left(-3 \lt X \lt 7\right)=P\left(X \lt 7\right)-P\left(X \lt -3\right)

P\left(-3 \lt X \lt 7\right)\approx 0.9\ 525-0.475

In order to find P\left(a \lt X \lt b\right) when X doesn't follow the standard normal distribution, we need to calculate the z -scores of a and b.

When \mu is the mean and \sigma is the standard deviation, the z -score is the number of standard deviations an observation lies from the mean:

• z_{a}=\dfrac{a-\mu}{\sigma}
• z_{b}=\dfrac{b-\mu}{\sigma}

In our problem, the mean is \mu=4 and the standard deviation is \sigma=7. We calculate the z -scores for -10 and 15 as:

• z_{-10}=\dfrac{-10-4}{7}=-2
• z_{15}=\dfrac{15-4}{7}\approx 1.57

The table of standard normal probabilities gives the probability of observing an outcome below the given z -score:

• P\left(X \lt a\right)\approx P\left(Z \lt z_{a}\right)
• P\left(X \lt b\right)\approx P\left(Z \lt z_{b}\right)

Looking at the standard normal probabilities table, we find P\left(Z \lt -2\right) and P\left(Z \lt 1.57\right).

• P\left(X \lt -10\right)\approx P\left(Z \lt -2\right)=1-P\left(Z \lt 2\right)=1-0.9\ 772=0.228
• P\left(X \lt 15\right)\approx P\left(Z \lt 1.57\right)=0.9\ 418

We know that:

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

Here we have:

P\left(-10 \lt X \lt 15\right)=P\left(X \lt 15\right)-P\left(X \lt -10\right)

P\left(-10 \lt X \lt 15\right)\approx 0.9\ 418-0.228

In order to find P\left(a \lt X \lt b\right) when X doesn't follow the standard normal distribution, we need to calculate the z -scores of a and b.

When \mu is the mean and \sigma is the standard deviation, the z -score is the number of standard deviations an observation lies from the mean:

• z_{a}=\dfrac{a-\mu}{\sigma}
• z_{b}=\dfrac{b-\mu}{\sigma}

In our problem, the mean is \mu=8 and the standard deviation is \sigma=3. We calculate the z -scores for 1 and 9 as:

• z_{1}=\dfrac{1-8}{3}\approx -2.33
• z_{9}=\dfrac{9-8}{3}\approx 0.33

The table of standard normal probabilities gives the probability of observing an outcome below the given z -score:

• P\left(X \lt a\right)\approx P\left(Z \lt z_{a}\right)
• P\left(X \lt b\right)\approx P\left(Z \lt z_{b}\right)

Looking at the standard normal probabilities table, we find P\left(Z \lt -2.33\right) and P\left(Z \lt 0.33\right).

• P\left(X \lt 1\right)\approx P\left(Z \lt -2.33\right)=1-P\left(Z \lt 2.33\right)\approx 1-0.9\ 901\approx 0.99
• P\left(X \lt 9\right)\approx P\left(Z \lt 0.33\right)\approx 0.6\ 293

We know that:

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

Here we have:

P\left(1 \lt X \lt 9\right)=P\left(X \lt 9\right)-P\left(X \lt 1\right)

P\left(1 \lt X \lt 9\right)\approx 0.6\ 293-0.99=0.6\ 194

In order to find P\left(a \lt X \lt b\right) when X doesn't follow the standard normal distribution, we need to calculate the z -scores of a and b.

When \mu is the mean and \sigma is the standard deviation, the z -score is the number of standard deviations an observation lies from the mean:

• z_{a}=\dfrac{a-\mu}{\sigma}
• z_{b}=\dfrac{b-\mu}{\sigma}

In our problem, the mean is \mu=5 and the standard deviation is \sigma=2. We calculate the z -scores for 3 and 7 as:

• z_{3}=\dfrac{3-5}{2}=-1
• z_{7}=\dfrac{7-5}{2}=1

The table of standard normal probabilities gives the probability of observing an outcome below the given z -score:

• P\left(X \lt a\right)\approx P\left(Z \lt z_{a}\right)
• P\left(X \lt b\right)\approx P\left(Z \lt z_{b}\right)

Looking at the standard normal probabilities table, we find P\left(Z \lt -1\right) and P\left(Z \lt 1\right).

• P\left(X \lt 3\right)\approx P\left(Z \lt -1\right)=1-P\left(Z \lt 1\right)\approx 1-0.8\ 413=0.1\ 587
• P\left(X \lt 7\right)\approx P\left(Z \lt 1\right)\approx 0.8\ 413

We know that:

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

Here we have:

P\left(3 \lt X \lt 7\right)=P\left(X \lt 7\right)-P\left(X \lt 3\right)

P\left(3 \lt X \lt 7\right)\approx 0.8\ 413-0.1\ 587=0.6\ 826

In order to find P\left(a \lt X \lt b\right) when X doesn't follow the standard normal distribution, we need to calculate the z -scores of a and b.

When \mu is the mean and \sigma is the standard deviation, the z -score is the number of standard deviations an observation lies from the mean:

• z_{a}=\dfrac{a-\mu}{\sigma}
• z_{b}=\dfrac{b-\mu}{\sigma}

In our problem, the mean is \mu=10 and the standard deviation is \sigma=3. We calculate the z -scores for 4 and 16 as:

• z_{4}=\dfrac{4-10}{3}=-2
• z_{16}=\dfrac{16-10}{3}=2

The table of standard normal probabilities gives the probability of observing an outcome below the given z -score:

• P\left(X \lt a\right)\approx P\left(Z \lt z_{a}\right)
• P\left(X \lt b\right)\approx P\left(Z \lt z_{b}\right)

Looking at the standard normal probabilities table, we find P\left(Z \lt -2\right) and P\left(Z \lt 2\right).

• P\left(X \lt 4\right)\approx P\left(Z \lt -2\right)=1-P\left(Z \lt 2\right)\approx 1-0.9\ 772=0.228
• P\left(X \lt 16\right)\approx P\left(Z \lt 2\right)\approx 0.9\ 772

We know that:

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

Here we have:

P\left(4 \lt X \lt 16\right)=P\left(X \lt 16\right)-P\left(X \lt 4\right)

P\left(4 \lt X \lt 16\right)\approx 0.9\ 772-0.228=0.9\ 544

In order to find P\left(a \lt X \lt b\right) when X doesn't follow the standard normal distribution, we need to calculate the z -scores of a and b.

When \mu is the mean and \sigma is the standard deviation, the z -score is the number of standard deviations an observation lies from the mean:

• z_{a}=\dfrac{a-\mu}{\sigma}
• z_{b}=\dfrac{b-\mu}{\sigma}

In our problem, the mean is \mu=6 and the standard deviation is \sigma=4. We calculate the z -scores for -3 and 7 as:

• z_{-3}=\dfrac{-3-6}{4}=-2.25
• z_{7}=\dfrac{7-6}{4}=0.25

The table of standard normal probabilities gives the probability of observing an outcome below the given z -score:

• P\left(X \lt a\right)\approx P\left(Z \lt z_{a}\right)
• P\left(X \lt b\right)\approx P\left(Z \lt z_{b}\right)

Looking at the standard normal probabilities table, we find P\left(Z \lt -2.25\right) and P\left(Z \lt 0.25\right).

• P\left(X \lt -3\right)\approx P\left(Z \lt -2.25\right)=1-P\left(Z \lt 2.25\right)\approx 1-0.9\ 878=0.122
• P\left(X \lt 7\right)\approx P\left(Z \lt 0.25\right)\approx 0.5\ 987

We know that:

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

Here we have:

P\left(-3 \lt X \lt 7\right)=P\left(X \lt 7\right)-P\left(X \lt -3\right)

P\left(-3 \lt X \lt 7\right)\approx 0.5\ 987-0.122=0.5\ 865