5.5 Working With Any Normal Distribution

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].

1. 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.

2. 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.

3. 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.

4. 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.

5. 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.

6. 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].