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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 ).

This is a normal distribution curve. A value, x, is labeled on the horizontal axis, X. A vertical line extends from point x to the curve, and the area under the curve to the left of x is shaded. The area of this shaded section represents the probability that a value of the variable is less than 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. There are instructions given as necessary for the TI-83+ and TI-84 calculators.

Note

To calculate the probability, use the probability tables provided in [link] without the use of technology. The tables include instructions for how to use them.

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

This is a normal distribution curve. The peak of the curve coincides with the point 63 on the horizontal axis. The point 65 is also labeled. A vertical line extends from point 65 to the curve. The probability area to the right of 65 is shaded; it is equal to 0.3446.

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

Go into 2nd DISTR .
After pressing 2nd DISTR , press 2:normalcdf .

The syntax for the instructions are as follows:

normalcdf(lower value, upper value, mean, standard deviation) For this problem: normalcdf(65,1E99,63,5) = 0.3446. You get 1E99 (= 10 99 ) by pressing 1 , the EE key (a 2nd key) and then 99 . Or, you can enter 10^99 instead. The number 10 99 is way out in the right tail of the normal curve. We are calculating the area between 65 and 10 99 . In some instances, the lower number of the area might be –1E99 (= –10 99 ). The number –10 99 is way out in the left tail of the normal curve.

Historical note

The TI probability program calculates a z -score and then the probability from the z -score. 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 = 65  – 63 5 = 0.4

Area to the left is 0.6554.

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

Calculate the z -score:

*Press 2nd Distr
*Press 3:invNorm (
*Enter the area to the left of z followed by )
*Press ENTER .
For this Example, the steps are
2nd Distr
3:invNorm (.6554) ENTER
The answer is 0.3999 which rounds to 0.4.

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.

normalcdf(0,85,63,5) = 1 (rounds to one)

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

This is a normal distribution curve. The peak of the curve coincides with the point 63 on the horizontal axis. A point, k, is labeled to the right of 63. A vertical line extends from k to the curve. The area under the curve to the left of k is shaded. This represents the probability that x is less than k: P(x < k) = 0.90

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. To get this answer on the calculator, follow this step:

Questions & Answers

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Missy Reply
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Joseph
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Preparation and Applications of Nanomaterial for Drug Delivery
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Damian
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Professor
I think
Professor
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LITNING Reply
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LITNING
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Source:  OpenStax, Introduction to statistics i - stat 213 - university of calgary - ver2015revb. OpenStax CNX. Oct 21, 2015 Download for free at http://legacy.cnx.org/content/col11874/1.3
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