# 5.5 Working With Any Normal Distribution

Through standardization, we can solve problems related to any normal distribution [latex]N(\mu, \sigma)[/latex] using the standard normal table. The following diagram shows the procedure.

Example: Working With Any Normal Distribution Using the Standard Normal Table (Table II)

Suppose the grades in a Statistics class are approximately normally distributed with a mean 70 and a standard deviation 10, i.e., [latex]X \sim N(\mu=70, \sigma =10)[/latex].

- Find the percentage of students whose grades are below 60.

[latex]P(X < 60) = P \left( \frac{X - \mu}{\sigma} < \frac{60-\mu}{\sigma} \right) = P\left(Z < \frac{60-70}{10}\right) = P(Z < -1) = 0.1587.[/latex]

15.87% of the students have a grade below 60.

- Find the percentage of students whose grades are above 95.

[latex]\begin{align*} P(X \: > \: 95) &= P\left(\frac{X-\mu}{\sigma} \: > \: \frac{95 - \mu}{\sigma}\right) \\ &= P\left(Z \: > \: \frac{95-70}{10}\right) \\ &= P(Z \: > \: 2.5) \\ &= 1 - P(Z < 2.5) \\ &= 1 - 0.9938 = 0.0062. \end{align*}[/latex]

or [latex]P(X \: > \: 95) = P(Z \: > \: 2.5) = P(Z < -2.5) = 0.0062[/latex].

Only 0.62% of students have a grade above 95.

- Find the percentage of students whose grades are between 60 and 90.The [latex]z[/latex]-score for 60 is [latex]z = \frac{x - \mu}{\sigma} = \frac{60-70}{10} = -1[/latex]; the [latex]z[/latex]-score for 90 is [latex]z = \frac{x - \mu}{\sigma} = \frac{90-70}{10} = 2[/latex]. [latex]\begin{align*} P(60 < X < 90) &= P(-1 < Z < 2) \\ &= P(Z < 2) - P(Z < -1) \\ &= 0.9772 - 0.1597 \\ &= 0.8185. \end{align*}[/latex]
81.85% of the students have a grade between 60 and 90.

- Suppose the bottom 5% of students fail. Find the minimum grade needed in order to pass the course.

Find the grade *x* that bounds the bottom 5% of grades, i.e., [latex]P(X \leq x) = 0.05[/latex]. The [latex]z[/latex]-score that captures the bottom 5% of the standard normal distribution is [latex]z=-1.645[/latex]. Therefore, the corresponding *x* value is

[latex]x = \mu +z \times \sigma = 70 + (-1.645) \times 10 = 70 - 16.45 = 53.55.[/latex]

The passing grade is 53.55.

- Suppose the top 2% of students will get an A. Find the minimum grade needed in order to obtain an A.

Find the grade *x* that bounds the top 2% of grades, i.e., [latex]P(X \ge x) = 0.02[/latex]. The [latex]z[/latex]-score that captures the top 2% of the standard normal distribution is [latex]z_{0.02}=2.05[/latex]. Therefore, the corresponding *x* value is

[latex]x = \mu +z \times \sigma = 70 + (2.05) \times 10 = 70 + 20.5 = 90.5.[/latex]

The cutoff is 90.5.

- Find the quartiles of student grades.

First, observe that the first, second and third quartiles of the standard normal distribution are [latex]z_1 = -0.67, z_2 = 0[/latex] and [latex]z_3=0.67[/latex], since [latex]P(Z < -0.67)\approx 0.25, P(Z<0)=0.5[/latex], and [latex]P(Z < 0.67)\approx 0.75[/latex]. Thus, the first, second and third quartiles of student grades are:

[latex]Q_1 = \mu + z_1 \times \sigma = 70 + (-0.67) \times 10 = 63.3,[/latex]

[latex]Q_2 = \mu + z_2 \times \sigma = 70 + (0) \times 10 = 70,[/latex]

[latex]Q_3 = \mu + z_3 \times \sigma = 70 + (0.67) \times 10 = 76.7.[/latex]

Thus, the quartiles of student grades are:

[latex]Q_1= 63.3, \quad Q_2 = 70, \quad Q_3 =76.7.[/latex]

Note: Recall that one of the properties of a symmetric distribution is that the mean and median are equal. So, we could have alternatively used the symmetry of a normal distribution to establish that [latex]Q_2 = \mu =70[/latex].

Exercise: Work With Any Normal Distribution Using Standard Normal Table (Table II)

Suppose the weight of boxes of a certain brand of cereal follows a normal distribution with a mean of 1,000 grams and a standard deviation of 40 grams.

- A box is rejected by the quality control department if its weight is below 950 grams. What percentage of boxes will be rejected?
- Find the percentage of boxes with weight in between 980 grams and 1,010 grams.
- Find the percentage of boxes with weight above 1,010 grams.
- Determine the 40th percentile for the weight of the boxes of cereal.
- A particular box of cereal is weighed, and it is determined that 5% of boxes are heavier than this particular box. What is the approximate weight of this box of cereal?

## Show/Hide Answer

- [latex]P(X < 950)=[/latex] [latex]P(Z < (950-1000)/40)=[/latex] [latex]P(Z < -1.25)=0.1056[/latex]. That is 10.56%.
- [latex]P(980 < X < 1010 )[/latex] [latex]= P (-0.5 < Z < 0.25)[/latex] [latex]= P(Z < 0.25) - P(Z < -0.5)[/latex] [latex]= 0.5987 - 0.3085 = 0.2902[/latex], that is, 29.02%.
- [latex]P(X \: > \: 1010) = P(Z \: > \: (1010 - 100)/40))[/latex] [latex]= P(Z \: > \: 0.25)[/latex] [latex]= 1 - P(Z < 0.25) = 1 - 0.5987 = 0.4013[/latex], that is, 40.13%.
- The 40th percentile of the standard normal distribution is [latex]z=-0.25[/latex]. Therefore, [latex]x = 1000 + (-0.25) \times 40 = 990[/latex], so that the 40th percentile among boxes of cereal is 990 grams.
- The 95th percentile of the standard normal distribution is [latex]z=1.645[/latex]. Therefore, [latex]x = 1000 + (1.645) \times 40 = 1065.8[/latex], so that the 95th percentile among boxes of cereal is 1065.8 grams.