5.7 Calculations of probabilities

Let $X=$ a score on the final exam. $X$ ~ $N(63,5)$ , where $\mu =63$ and $\sigma =5$

Draw a graph.

Then, find $P(x> 65)$ .

$P(x> 65)=0.3446$ (calculator or computer)

The probability that one student scores more than 65 is 0.3446.

Using the TI-83+ or the TI-84 calculators, the calculation is as follows. Go into 2nd DISTR .

After pressing 2nd DISTR , press 2:normalcdf .

The syntax for the instructions are shown below.

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 areabetween 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.

$z=\frac{65-63}{5}=0.4$ . Area to the left is 0.6554. $P(x> 65)=P(z> 0.4)=1-0.6554=0.3446$

Draw a graph.

Then find $P(x< 85)$ . Shade the graph. $P(x< 85)=1$ (calculator or computer)

The probability that one student scores less than 85 is approximately 1 (or 100%).

The TI-instructions and answer are as follows:

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

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

Let $k$ = the 90th percentile. $k$ is located on the x-axis. $P(x< k)$ is the area to the left of $k$ . The 90th 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 10% are the same or higher. $k$ is often called a critical value .

$k=69.4$ (calculator or computer)

The 90th percentile is 69.4. This means that 90% of the test scores fall at or below 69.4 and 10% fall at or above.For the TI-83+ or TI-84 calculators, use invNorm in 2nd DISTR . invNorm(area to the left, mean, standard deviation)For this problem, invNorm(0.90,63,5) = 69.4

Find the 70th percentile.

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

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

invNorm(0.70,63,5) = 65.6