31 6.2 Using the Normal Distribution
The blue shaded area in the following graph indicates the area to the left of x.
P(X < a) = Area to the left of the vertical line through a. (blue area)
P(X > a) = Area to the right of the vertical line through a. (white area)
P(X > a) is also equal to 1 – P(X < a).
Remember,
P(X < a) = P(X ≤ a) and P(X > a) = P(X ≥ a) 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.
Additionally, this link houses a tool that allows you to explore the normal distribution with varying means and standard deviations as well as associated probabilities. The following video explains how to use the tool.
Example 1
If the area to the left is 0.0228, what is the area to the right?
Show Answer
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?
[practice-area rows=”2″][/practice-area]
Show Answer
1 − 0.012 = 0.988
Example 2 (by using TI-83/84 Calculators)
The final exam scores in a statistics class were normally distributed with a mean of 63 and a standard deviation of 5.
- Find the probability that a randomly selected student scored more than 65 on the exam.
- Find the probability that a randomly selected student scored less than 85.
- Find the 90th percentile (that is, find the score k that has 90% of the scores below k and 10% of the scores above k).
- Find the 70th percentile (that is, find the score k such that 70% of scores are below k and 30% of the scores are above k).
Solution
- Find the probability that a randomly selected student scored more than 65 on the exam.Let X = a score on the final exam. X ~ N(63, 5), where mean, μ = 63 and standard deviation,σ = 5.
- Draw a graph. Then, find P(x > 65).
- Using TI-Calculator:
Go into2nd DISTR
. Then press2:normalcdf
.
The syntax for the instructions are as follows: normalcdf (lower value, upper value, mean, standard deviation).
For this problem: normalcdf (65,10^99,63,5) and press “=”.
The answer is 0.3446. - Therefore, P(x > 65) = 0.3446.
The probability that any student selected at random scores more than 65 is 0.3446.
******************************************************************************************************************Extra Note:
The number 1099 is way out in the right tail of the normal curve. We are calculating the area when final exam score is between between 65 and 1099.
In some instances, the lower number of the area might be –1099.
The number –1099 is way out in the left tail of the normal curve.
[latex]\displaystyle{z}=\frac{{{65}-{63}}}{{5}}={0.4}[/latex]
We can solve that area to the left of z-score 0.4 is 0.6554 by inputting normalcdf(-10^99, 65, 63, 5) to TI-Calculator.
Hence, P(x > 65) = P(z > 0.4) = 1 – 0.6554 = 0.3446***************************************************************************************************************************
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