House of Math

The probability of getting a result between $a$ and $b$ is given by the area bounded by the graph, the $x$ -axis and vertical lines at $x = a$ and $x = b$ . $P (a \geq x \geq b)$ can be found in a probability table or through the use of digital tools.

When you’re working with a normal distribution by hand, you need to convert the normal distribution into the standard normal distribution. Then, you can use the standardized table that contains all the solutions. A standardized normal distribution is a normal distribution with an expected value equal to 0 and a standard deviation equal to 1. When converting into the standard normal distribution, you use the following conversion formula:

Formula

Conversion to Standard Normal Distribution

Z = \frac{X - μ}{σ}

where the random variable $Z$ is distributed just like the standard normal distribution.

Example 1

The birth weight of a newborn girl is considered normally distributed with an expectation value of $3.50 kg$ and a standard deviation of $0.48 kg$ .

1.: What is the probability that a random girl weighs less than $2.5 kg$ when born?
2.: What is the probability that a random girl weighs more than $4.0 kg$ when born?
3.: What is the probability that a random girl weighs between $2.5 kg$ and $4.0 kg$ when born?

1.

You want to find

P (X \leq 2.5)

. The calculation looks like this:

\begin{array}{l} P (X \leq 2.5) & = P (\frac{X - μ}{σ} \leq \frac{2.5 - 3.5}{0.48}) \\ = P (Z \leq - 2.0833) \end{array}

You look up the $Z$ -value in the probability table for the normal distribution and find that

\begin{array}{l} P (Z \leq - 2.0833) \Leftrightarrow Z & = 0.0188 \\ = 1.88 % \end{array}

\begin{array}{l} P (Z \leq - 2.0833) \Leftrightarrow Z = 0.0188 = 1.88 % \end{array}

The probability of a new born girl weighing less than

2.5

kg is

1.88

%.

2.

You want to find

P (X \geq 4.0)

. The calculation looks like this:

\begin{array}{l} P (X \geq 4.0) & = 1 - P (X \leq 4.0) \\ = 1 - P (\frac{X - μ}{σ} \leq \frac{4.0 - 3.5}{0.48}) \\ = 1 - P (Z \leq 1.04) \end{array}

You look up the $Z$ -value in the table and find that

P (Z \leq 1.04) = 0.8508

which gives you that

\begin{array}{l} 1 - P (Z \leq 1.04) & = 1 - 0.8508 \\ = 0.1492 \\ = 14.92 % \end{array}

The probability that a random newborn girl weighs more than $4.0$ kg is $14.92$ %.

3.

You want to find

P (2.5 \leq X \leq 4.0)

. In this case, you have to set up a double inequality. It looks like this:

\begin{array}{l} P (2.5 \leq X \leq 4.0) & = P (\frac{2.5 - 3.5}{0.48} \\ \leq \frac{X - μ}{σ} \\ \leq \frac{4.0 - 3.5}{0.48}) \\ = P (- 2.08 \leq Z \\ \leq 1.04) \\ = P (Z \leq 1.04) \\ - P (Z \leq - 2.08) \end{array}

\begin{array}{l} P (2.5 \leq X \leq 4.0) & = P (\frac{2.5 - 3.5}{0.48} \leq \frac{X - μ}{σ} \leq \frac{4.0 - 3.5}{0.48}) \\ = P (- 2.08 \leq Z \leq 1.04) \\ = P (Z \leq 1.04) - P (Z \leq - 2.08) \end{array}

From the table you find the values of the respective

Z

-values, and their difference is

\begin{array}{l} P (2.5 \leq X \leq 4.0) & = 0.8508 - 0.0188 \\ = 0.8320 \\ = 83.2 % \end{array}

The probability that a random newborn girl weighs between $2.5$ kg and $4.0$ kg is $83.2$ %.

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