Suppose that grade point averages of undergraduate students at one university have a bell-shaped distribution with a mean of 2.51 and a standard deviation of 0.41 . Using the empirical rule, what percentage of the students have grade point averages that are less than 1.69 ? Please do not round your answer.

Approximately [tex]68.27\%[/tex] of the observations would be within [tex]1\, \sigma[/tex] from the mean, in the interval [tex](\mu - \sigma,\, \mu + \sigma)[/tex].
Approximately [tex]95.45\%[/tex] of the observations would be within [tex]2\, \sigma[/tex] from the mean, in the interval [tex](\mu - 2\, \sigma,\, \mu + 2\, \sigma)[/tex].
Approximately [tex]99.73\%[/tex] of the observations would be within [tex]3\, \sigma[/tex] from the mean, in the interval [tex](\mu - 3\, \sigma,\, \mu + 3\, \sigma)[/tex].

In the distribution in this question, it is given that [tex]\mu = 2.51[/tex] and [tex]\sigma = 0.41[/tex]. Hence:

Approximately [tex]68.27\%[/tex] of the observations would be in the interval [tex](2.10,\, 2.92)[/tex].
Approximately [tex]95.45\%[/tex] of the observations would be in the interval [tex](1.69,\, 3.33)[/tex].
Approximately [tex]99.73\%[/tex] of the observations would be in the interval [tex](1.28,\, 3.74)[/tex].

This question is asking for the probability [tex]P(X < 1.69)[/tex].

By the empirical rule, [tex]P(1.69 < X < 3.33) \approx 95.45\%[/tex]. Because the distribution is symmetric, and the interval [tex](1.69,\, 3.33)[/tex] is centered at the mean of the distribution, the following probabilities would be equal:

[tex]P(X < 1.69)[/tex], to the left of the interval, and
[tex]P(X > 3.33)[/tex], to the right of the interval.

Additionally, the sum of the probability over the three intervals should be [tex]1[/tex]. Therefore:

[tex]P(X < 1.69) + P(1.69 < X < 3.33) + P(X > 3.33) = 1[/tex].

Since [tex]P(X < 1.69) = P(X > 3.33)[/tex], the equation becomes:

[tex]2\, P(X < 1.69) + P(1.69 < X < 3.33) = 1[/tex].

[tex]\begin{aligned} P(X < 1.69) &= \frac{1 - P(1.69 < X < 3.33)}{2} \\ &\approx \frac{100\% - 95.45\%}{2} \\ &= 2.275\%\end{aligned}[/tex].

In other words, approximately [tex]2.275\%[/tex] of the observations would be less than [tex]1.69[/tex].