# 6.2 Using the normal distribution

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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. 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

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 = = 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

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

how do you find z if you only know the area of .0808
Cady Reply
How to take a random sample of 30 observations
Hamna Reply
you can use the random function to generate 30 numbers or observation
smita
How we can calculate chi-square if observed x٫y٫z/frequency 40,30,20 Total/90
Insha Reply
calculate chi-square if observed x,y,z frequency 40,30,20total 90
Insha
find t value,if boysN1, ،32,M1,87.43 S1square,39.40.GirlsN2,34,M2,82.58S2square,40.80 Determine whether the results are significant or insignificant
Insha
The heights of a random sample of 100 entering HRM Freshman of a certain college is 157 cm with a standard deviation of 8cm. test the data against the claim that the overall height of all entering HRM students is 160 cm. previous studies showed that
Crispen Reply
complete the question.. as data given N = 100,mean= 157 cm, std dev = 8 cm..
smita
Z=x-mu/ std dev
smita
find the mean of 25,26,23,25,45,45,58,58,50,25
Asmat Reply
add all n divide by 10 i.e 38
smita
38
hhaa
amit
1 . The “average increase” for all NASDAQ stocks is the:
Jamshaid Reply
STATISTICS IN PRACTICE: This is a group assignment that seeks to reveal students understanding of statistics in general and it’s practical usefulness. The following are the guidelines; 1.      Each group has to identify a natural process or activity and gather data about/from the process. 2.
Kofi Reply
The diameter of an electric cable,say, X is assumed to be continoues random variable with p.d.f f(x)=6x(1-x); ≤x≤1 a)check that f(X) is p.d.f b) determine a number b such that p(Xb)
Syed Reply
A manufacturer estimate 3% of his output is defective. Find the probability that in a sample of 10 items (a) less than two will be defective (b) more than two will be defective.
ISAIAH Reply
A manufacturer estimates that 3% of his output of a small item is defective. Find the probabilities that in a sample of 10 items (a) less than two and (b) more than two items will be defective.
ISAIAH
use binomial distribution with parameter n=10, p= 0.03, q=0.97
Shivprasad
the standard deviation of a symmetrical distribution is 7.8 . what must be the value of forth moment about the mean in order that distribution be a) leptokurtic b) mesokurtic c) platy kyrtic intrept the obtain value of a b and c
Tushar Reply
A researcher observed that four out of every ten of their products are normally defective. A total of 360 samples of the products were being tested. If the sample is normally distributed and 220 of the products were identified to be faulty, test the hypothesis that the observation of the res
Adepoju Reply
false
HariPrasad
please answer the ques"following values are obtained from life table T15=3,493,601 and e°15=44.6 then expected number of person alive at exact age 15 will be "
vinay
make it clear
Kagimu
how x minus x bar is equal to zero
Kashif Reply
When the mean (X bar) of the sample and the datapoint-in-context (X) from the same sample are the same, then it (X minus X bar) is equal to 0
Johns
e.g. mean of. sample is 3 and one of the datapoints in that sample is also 3
Johns
a numerical value used as a summary measure for a sample such as a sample mean is known as
rana Reply
differentiate between qualitative and quantitative variables
rana Reply
qualitative variables are descriptive while quantitative are numeric variables
Chisomo
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Yunisa Reply
Dear Yunisa there are different formulas used in statistics depending on wnat you want to measure. It would be helpful if you can be more specific
LAMIN

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