Question

If the scores for a test have a mean of 70 and a standard deviation of 12. the test scores are known to be approximately normal. find the percentage of scores that will fall below 50

Answer 1

The scores have a normal distribution mean 70 sigma 12. We seek the area under the bell curve from negative infinity to 50, which gives us the probability of a score in that range.

We normalize the score into a z score equal to the number of standard deviations from the mean; then we can use the standard normal table to get the area from negative infinity to z.

$z = \dfrac{50 -70}{12} = -\dfrac 5 3 \approx -1.67$

From the 68-95-99.7 rule we know that's less than 16% and more than 2.5%; we look it up in the unit normal table and get

$p = .0475 = 4.75\%$

We expect 4.75% of the scores to be below 50.

Answer 2

Let X be the score of the test. X follows normal distribution with mean μ = 70 and standard deviation σ =12

The percentage of scores that will fall below 50 is

P(X < 50) = $P(\frac{x - mean}{standard deviation} < \frac{50 - 70}{12}$

= P(z < -1.67)

Using standard normal probability table to find probability below -1.67

P(X < 50) = P(Z < -1.67) = 0.0475

The probability that scores will fall below 50 is 0.0475

Converting probability into percentage

0.0475 * 100 = 4.75

The percentage of score will fall below 50 is 4.75%