# 5.2 Using the normal distribution -- rrc math1020

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The shaded area in the following graph indicates the area to the left of x . This area is represented by the probability P ( X < x ). Normal tables, computers, and calculators provide or calculate the probability P ( X < x ).

The area to the right is then P ( X > x ) = 1 – P ( X < x ). Remember, P ( X < x ) = Area to the left of the vertical line through x . P ( X < x ) = 1 – P ( X < x ) = Area to the right of the vertical line through x . P ( X < x ) is the same as P ( X x ) and P ( X > x ) is the same as P ( X x ) for continuous distributions.

## Calculations of probabilities

Probabilities are calculated using technology.

## Note

To calculate the probability, you can use probability tables without the use of technology.

If the area to the left is 0.0228, then the area to the right is 1 – 0.0228 = 0.9772.

## Try it

If the area to the left of x is 0.012, then what is the area to the right?

1 − 0.012 = 0.988

The final exam scores in a statistics class were normally distributed with a mean of 63 and a standard deviation of five.

a. Find the probability that a randomly selected student scored more than 65 on the exam.

a. Let X = a score on the final exam. X ~ N (63, 5), where μ = 63 and σ = 5

Draw a graph.

Then, find P ( x >65).

P ( x >65) = 0.3446

The probability that any student selected at random scores more than 65 is 0.3446.

## Historical note

Before technology, the z -score was looked up in a standard normal probability table (because the math involved is too cumbersome) to find the probability. In this example, a standard normal table with area to the left of the z -score was used. You calculate the z -score and look up the area to the left. The probability is the area to the right.

z = = 0.4

Area to the left is 0.6554.

P ( x >65) = P ( z >0.4) = 1 – 0.6554 = 0.3446

b. Find the probability that a randomly selected student scored less than 85.

b. Draw a graph.

Then find P ( x <85), and shade the graph.

Using a computer or calculator, find P ( x <85) = 1.

The probability that one student scores less than 85 is approximately one (or 100%).

c. Find the 90 th percentile (that is, find the score k that has 90% of the scores below k and 10% of the scores above k ).

c. Find the 90 th percentile. For each problem or part of a problem, draw a new graph. Draw the x -axis. Shade the area that corresponds to the 90 th percentile.

Let k = the 90 th percentile. The variable k is located on the x -axis. P ( x < k ) is the area to the left of k . The 90 th percentile k separates the exam scores into those that are the same or lower than k and those that are the same or higher. Ninety percent of the test scores are the same or lower than k , and ten percent are the same or higher. The variable k is often called a critical value .

k = 69.4

The 90 th percentile is 69.4. This means that 90% of the test scores fall at or below 69.4 and 10% fall at or above.

d. Find the 70 th percentile (that is, find the score k such that 70% of scores are below k and 30% of the scores are above k ).

d. Find the 70 th percentile.

Draw a new graph and label it appropriately. k = 65.6

The 70 th percentile is 65.6. This means that 70% of the test scores fall at or below 65.5 and 30% fall at or above.

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