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STATISTICS EXPLORATION # 6
THE NORMAL AND OTHER CONTINUOUS PROBABILITY DISTRIBUTIONS
PURPOSE - to use MINITAB to
BACKGROUND INFORMATION
PROCEDURES
First, load the MINITAB (windows version) software as described in Exploration #0.
NOTE: Several continuous probability distributions can be investigated through the procedures included MINITAB, however, this exploration will concentrate only on the normal probability distribution.
NOTE: The procedures presented in these explorations may not be the only way to achieve the end results. Also, whenever graphs are presented, only the MINITAB graphics features will be used.
This section will illustrate how MINITAB can be used to generate random samples from a normal distribution and use the generated values to graph the corresponding normal curve.
Example 1: Generate a random sample of size 1000 from a standard normal distribution. Use the generated values to compute the probability density values for the generated values. Plot along the density values along the y-axis and the generated values from the normal distribution along the x-axis. Use the connect option to display the graph.
Recall: The standard normal distribution has a mean m =0 and s = 1.
First to generate the normal values, select Calc® Random Data. Figure 6.1 shows a partial display of the action.
Figure 6.1: Display of the result of selecting Calc® Random Data
To generate random data from the normal distribution, click on the Normal distribution in the Sample From Columns option. In the resulting dialog box, enter the values as shown in Figure 6.2. Click on the OK button and the data will be generated from the standard normal distribution and saved in column C1 (NORMAL).
Figure 6.2: Normal Distribution dialog box with Entries to simulate 1000 values from a Standard Normal Distribution
Next, we need to compute the Probability density for these simulated values.
To achieve this, select Calc® Probability Distributions® Normal. Fill in the entries as shown in Figure 6.3.
Note: Column C1 was renamed as NORMAL and column C2 was renamed as DENSITY.
Observe that we have selected the Probability density option. Also, observe that we retained the mean of 1 and standard deviation of 1 for the standard normal distribution. Since we needed the density values for the data in column C1, we use this as the Input column entry and use C2 as the Optional storage column for the density values.
Figure 6.3: Normal Distribution dialog box with Entries to compute the Probability Densities for the 1000 simulated values from a Standard Normal Distribution
Click on the OK button and the densities will be generated.
Next, we will plot the density values along the y-axis and the normal values in C1 along the x-axis. To achieve this, select Graph® Plot and in the Plot dialog box, fill in as shown in Figure 6.4. Note that we selected Connect as the Display option for the graph.
Figure 6.4: The Plot dialog box with entries to graph the DENSITY values< along the y-axis and the NORMAL values along the x-axis
Click on the OK button and the graph will be generated. Figure 6.5 shows the resulting graph.
Figure 6.5: The Standard Normal Probability distribution graph
Observations from the graph in Figure 6.5:
Figure 6.6 shows some of these properties.
Figure 6.6: The Standard Normal Probability distribution graph Displaying some of its Properties
In this section we will use MINITAB to help in computing probabilities for a normal random variable.
Example 2: Use MINITAB to determine P(Z > 1.0).
Note: Z is a standard normal random variable. That is, Z is normally distributed with a mean m = 0 and a standard deviation s = 1.
Figure 6.7 shows the area that represents P(Z > 1.0).
Figure 6.7: Shaded area represents P(Z > 1.0)
Now, for us to use MINITAB to help find this area, select Calc® Probability Distributions® Normal. The Normal Distribution dialog box will appear.
Select Cumulative probability.
Also, recall that Z is a standard normal random variable so the mean will be 0 and the standard deviation will be 1. Next select the Input constant option and type 1 in the text box. All of these are shown in Figure 6.8.
Figure 6.8: Normal Distribution dialog box with the Cumulative Probability selections
Click on the OK button and the cumulative probability will be computed and displayed in the Session window. This value is shown in Figure 6.9.
Figure 6.9: Display of the results for the Cumulative Probability
Now, from Figure 6.9, P(X < = x) = P(X <= 1.0) = 0.8413 represents the value of P(Z £ 1.0). That is, P(Z £ 1) = 0.8413. This shown in Figure 6.10.
NOTE: Since Z is a continuous random variable, P(Z £ 1.0) = P(Z < 1.0).
Figure 6.10: Area representing P(Z £ 1.0)
Recall, the original problem was to find P(Z > 1). Now, since the total probability under the curve and the x-axis is 1, then we can see that P(Z > 1) = 1 ? P(Z £ 1) = 1 ? 0.8413 = 0.1587.
Example 3: Use MINITAB to determine P(Z > -2.2).
Figure 6.11 shows the area that represents P(Z > -2.2).
Figure 6.11: Shaded area represents P(Z > -2.2)
Now, for us to use MINITAB to help find this area, select Calc® Probability Distributions® Normal. The Normal Distribution dialog box will appear.
Again select Cumulative probability with the Input constant being ?2.2. Refer to Figure 6.8.
Click on the OK button and the cumulative probability will be computed and displayed in the Session window. This value is shown in Figure 6.12 and it is equal to P(Z £ -2.2).
Figure 6.12: Display of the results for P(Z £ -2.2)
Now, from Figure 6.12, P(X < = x) = P(X <= -2.2) = 0.0139 represents the value of P(Z £ -2.2). That is, P(Z £ -2.2) = 0.0139.
Recall, the original problem was to find P(Z > -2.2). Now, since the total probability under the curve and the x-axis is 1, then we can see that P(Z > -2.2) = 1 ? P(Z £ -2.2) = 1 ? 0.0139 = 0.9861.
Example 4: Use MINITAB to determine P(Z £ 1.8).
Figure 6.13 shows the area that represents P(Z £ 1.8).
Figure 6.13: Shaded area represents P(Z £ 1.8)
Now, for us to use MINITAB to help find this area, select Calc® Probability Distributions® Normal. The Normal Distribution dialog box will appear.
Again select Cumulative probability with the Input constant being 1.8. Refer to Figure 6.8.
Click on the OK button and the cumulative probability will be computed and displayed in the Session window. This value is shown in Figure 6.14 and it is equal to P(Z £ 1.8).
Figure 6.14: Display of the results for P(Z £ 1.8)
Now, from Figure 6.14, P(X < = x) = P(X <= 1.8) = 0.9641 represents the value of P(Z £ 1.8). That is, P(Z £ 1.8) = 0.9641.
Example 5: Use MINITAB to determine P(-2.7 £ Z £ 1.3).
Observe that we can write P(-2.7 £ Z £ 1.3) = P(Z £ 1.3) - P(Z £ -2.7), by using cumulative probabilities. This depicted below.
Figure 6.15 shows the area that represents P(-2.7 £ Z £ 1.3).
Figure 6.15: Shaded area represents P(-2.7 £ Z £ 1.3)
Now, for us to use MINITAB to help find this area, select Calc® Probability Distributions® Normal. The Normal Distribution dialog box will appear.
Again select Cumulative probability with the Input constant being 1.3. Repeat for an Input constant of ?2.7. Refer to Figure 6.16.
Click on the OK button and the cumulative probabilities will be computed and displayed in the Session window. These values are shown in Figure 6.16.
Figure 6.16: Display of the results for P(Z £ 1.3) and P(Z £ -2.7)
So, from Figure 6.16, P(-2.7 £ Z £ 1.3) = P(Z £ 1.3) - P(Z £ -2.7) = 0.9032 ? 0.0035 = 0.8997.
Example 6: A normal random variable X has a mean m = 100 and a standard deviation s = 2. Use MINITAB to determine the probability that X is greater than 95.
NOTE: MINITAB will compute probabilities for any normal distribution. It is not restricted to only the standard normal distribution.
Figure 6.17 shows the area that represents P(X > 95).
Figure 6.17: Shaded area represents P(X > 95)
Now, for us to use MINITAB to help find this area, select Calc® Probability Distributions® Normal. The Normal Distribution dialog box will appear.
Select Cumulative probability with the Input constant being 95. Make sure that you enter a mean of 100 and a standard deviation of 2 in the Mean and Standard deviation text boxes. You will need to do this since X is normal with a mean of 100 and a standard deviation of 2. Refer to Figure 6.18.
Figure 6.18: Normal Distribution dialog box with the Cumulative Probability selections
Click on the OK button and the cumulative probabilities will be computed and displayed in the Session window. These values are shown in Figure 6.19.
Figure 6.19: Display of the results for P(X £ 95)
Now, since P(X > 95) = 1 ? P(X £ 95). Then, from Figure 6.19,
P(X > 95) = 1 ? P(X £ 95) = 1 ? 0.0062 = 0.9938.
Example 7: A normal random variable X has a mean m = 100 and a standard deviation s = 2. Use MINITAB to determine the probability that X is less than 102.
NOTE: MINITAB will compute probabilities for any normal distribution. It is not restricted to only the standard normal distribution.
Figure 6.20 shows the area that represents P(X < 102).
Figure 6.20: Shaded area represents P(X < 102)
Again, for us to use MINITAB to help find this area, select Calc® Probability Distributions® Normal. The Normal Distribution dialog box will appear.
Select Cumulative probability with the Input constant being 102. Make sure that you enter a mean of 100 and a standard deviation of 2 in the Mean and Standard deviation text boxes. Again, you will need to do this since X is normal with a mean of 100 and a standard deviation of 2. Refer to Figure 6.18 except now the input constant will be 102.
Click on the OK button and the cumulative probabilities will be computed and displayed in the Session window. These values are shown in Figure 6.21.
Figure 6.21: Display of the results for P(X < 102)
From Figure 6.21, P(X < 102) = 0.8413.
Example 8: A normal random variable X has a mean m = 100 and a standard deviation s = 2. Use MINITAB to determine the probability that X lies between 94 and 104.
Observe that we can write P(94 £ X £ 104) = PX £ 104) - P(X£ 94), by using cumulative probabilities. This depicted below.
Figure 6.22 shows the area that represents P(94 £ X £ 104) .
Figure 6.22: Shaded area represents P(94 £ X £ 104)
Now, for us to use MINITAB to help find this area, select Calc® Probability Distributions® Normal. The Normal Distribution dialog box will appear.
Again select Cumulative probability with the Input constant being 104. Repeat for an Input constant of 94. Make sure that you enter a mean of 100 and a standard deviation of 2 in the Mean and Standard deviation text boxes. Again, you will need to do this since X is normal with a mean of 100 and a standard deviation of 2.
Click on the OK button and the cumulative probabilities will be computed and displayed in the Session window. These values are shown in Figure 6.23.
Figure 6.23: Display of the results for P(X £ 104) and P(X £ 94)
So, from Figure 6.23, P(94 £ X £ 104) = P(X £ 104) - P(X £ 94) = 0.9772 ? 0.0013 = 0.9759.
In the previous section we found probabilities associated with values of normal random variables. In this section, we will use MINITAB to help find values of any normal random variable when we are given the associated probabilities.
NOTE: The feature in MINITAB that allows us to find corresponding values for a random variable in a probability distribution is the Inverse cumulative probability option in the distribution.
Example 9: Assume that IQ scores have a normal distribution with a mean m = 100 and a standard deviation s = 15. Use MINITAB to determine the 75^{th} percentile of the IQ distribution.
Let the random variable X represent the IQ score. Let x_{0} represent the IQ score we are required to determine.
Now, since x_{0} represents the 75^{th} percentile for the IQ scores, it means that this value of x_{0} will be larger than 75% of all the IQ scores (when ordered from smallest to largest). Symbolically, we can write P(X £ x_{0}) = 0.75.
Figure 6.24 shows this relationship.
Figure 6.24: Shaded area represents P(X £ x_{0}) = 0.75
For us to use MINITAB to help find the value for x_{0}, select Calc® Probability Distributions® Normal. The Normal Distribution dialog box will appear.
Select Inverse cumulative probability with the Input constant being 0.75. Make sure that you enter a mean of 100 and a standard deviation of 15 in the Mean and Standard deviation text boxes respectively. You will need to do this since X, the IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. Figure 6.25 shows the dialog box with the appropriate entries.
Figure 6.25: Dialog box with entries to help determine x_{0} in the relationship P(X £ x_{0}) = 0.75
Click on the OK button and the inverse cumulative probability value will be computed and displayed in the Session window. This values is shown in Figure 6.26.
Figure 6.26: Session window output with value for x_{0} in the relationship P(X £ x_{0}) = 0.75
From Figure 6.26, P(X £ x_{0}) = 0.75 implies that x_{0} = 110.1173.
That is, the 75^{th} percentile for the IQ scores is approximately equal to 110.
Example 10: Assume that IQ scores have a normal distribution with a mean m = 100 and a standard deviation s = 15. Use MINITAB to determine the range for the middle 80% of the IQ scores.
Let the random variable X represent the IQ score. Let x_{1} represent the lower end point for the interval of IQ scores and let x_{2} represent the upper end point for the interval of IQ scores.
Now, since we are dealing with the middle 80% of the IQ scores, symbolically, we can write P(x_{1} £ X £ x_{2}) = 0.80.
Figure 6.27 shows this relationship.
Figure 6.27: Shaded area represents P(x_{1} £ X £ x_{2}) = 0.80
Now, since the area between x_{1} and x_{2} is 0.8, then the total area of the un-shaded portions in Figure 6.27 is 0.2. From the symmetry of the normal distribution, the area to the left of x_{1} will be equal to the area to the right of x_{2}. These areas in the tails of the distribution will be each equal to 0.1. This is depicted in Figure 6.28.
Figure 6.28: Display with all areas marked
From Figure 6.28 observe that the total area to the left of x_{2} will be equal to 0.9. Thus we can use the Inverse cumulative probability feature to obtain x_{2}. Select Calc® Probability Distributions ® Normal. The Normal Distribution dialog box will appear. Select Inverse cumulative probability with the Input constant being 0.90. Make sure that you enter a mean of 100 and a standard deviation of 15 in the Mean and Standard deviation text boxes respectively. You will need to do this since X, the IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. Refer to Figure 6.25 for an example of the dialog box.
Click on the OK button and the inverse cumulative probability value will be computed and displayed in the Session window. The value of 119.2233 which corresponds to the value of x_{2} is shown in Figure 6.29.
Figure 6.29: Session window output with value for x_{2} in the relationship P(X £ x_{2}) = 0.9
To obtain the value for x_{1} repeat with the input constant being equal to 0.1. The value of 80.7767 which corresponds to the value of x_{1} is shown in Figure 6.30.
Figure 6.30: Session window output with value for x_{1} in the relationship P(X £ x_{1}) = 0.1
Thus the range for the middle 80% of the IQ scores is obtained by subtracting the value of x_{1} from the value of x_{2}. That is the range = 119.2233 ? 80.7767 = 38.4466 or approximately a range of 39.
There are several techniques that can be used to establish whether a sample set of data values came from a normal distribution. These include the use of
The first four graphical techniques have been discussed in previous explorations. Here we will discuss normal probability plots.
Recall, a normal probability plot for a set of data is a scatter plot with the ranked data values along the vertical axis and the corresponding z-values from a standard normal distribution along the horizontal axis.
Also, when the data are normally distributed, a linear (straight line) trend will result in the normal probability plot. A nonlinear trend in the normal probability plot will suggest that the data are non-normal.
Example 11: Open the worksheet called Trees. This data file is one of the many data files that are packaged with the MINITAB software.
When the worksheet is opened, there will be data values listed for the three variables: Diameter, Height, and Volume.
Use MINITB to construct a normal probability for the variable Height.
To produce a normal probability plot for the variable Height, select Graph® Probability Plot. In the resulting dialog box, select Height for the Variables text box and select Normal as the Distribution. These selections are shown in Figure 31.
NOTE: Under the Options button in the dialog box, deselect the following two features: Display table of percentile estimates and Include confidence intervals in plot.
Figure 6.31: Probability Plot dialog box with Entries to Construct a Normal Probability Plot for the variable Height
Click on the OK button and the normal probability plot will be generated as shown in Figure 6.32.
Figure 6.32: Normal Probability Plot for the variable Height
Observe that the plot exhibit a linear trend. This would suggest that the sample values of the tree heights came from a normal distribution.
Figure 6.33 shows the normal probability plot for the diameter. Observe that the plot does not follow a straight-line pattern. This would indicate that the sample values for the tree volumes did not come from a normal distribution.
Figure 6.33: Normal Probability Plot for the variable Volume
In this section, we will investigate the normal approximation for the binomial distribution.
Recall, if X is a binomial random variable with n trials and probability of success p, then the
Suppose that Y is a normal random variable with mean and standard deviation .
We will use MINITAB to demonstrate this approximation of the binomial random variable with the normal random variable through the following example.
Example 12: A 1995 report from a Harvard University study claims that 34.9% of people in the age group of 25 to 29 years of age were homeowners. Suppose that a random sample of 100 from this age group is polled
Here X is a binomial random variable since:
Let the normal random variable Y have the same mean and standard deviation.
Note:
Although s is not greater than 5, the approximation should hold well.
We will use MINITAB to simulate 500 values from a binomial distribution with the mean and standard deviation computed above. Place headings of x-value, P(x-value), P(y-value) in columns C1, C2, and C3 respectively.
Refer to Exploration 5 to see how to simulate values and probabilities from the binomial distributions.
To compute the P(y-value) in column C3, use the 500 binomial values in C1 as the input column and use MINITAB to compute the densities for these values as normal values.
NOTE: In both simulations we are using the same mean and standard deviation.
Next, plot the P(x-value) and P(y-value) along the vertical axis and the x-value along the horizontal axis. You will need to superimpose both graphs on the set of axes.
For one of the graphs you may use a symbol to plot the points and for the other graph you may want to connect the plotted points. This will enable to see the two different plots.
To superimpose both graphs on the same axes, in the Plot dialog box, select the option of Frame and choose Multiple Graphs. In the Multiple Graphs dialog box select Overlay graphs on the same page. An example of the Multiple Graphs dialog box is shown in Figure 6.34.
Figure 6.34: Multiple Graphs dialog box
A sample plot of the superimposed graphs is shown in Figure 6.35.
Figure 6.35: Superimposed Binomial and Normal approximation
Observe from Figure 6.35 that the curves almost coincide. Thus one can conclude that the normal distribution is approximating the binomial probabilities very well.
NOTES
EXPLORATION #6: HOMEWORK ASSIGNMENT
Name: _____________________ Date: ______________________
Course #: ___________________ Instructor: _________________
P(X > 121) = ______________________________
P(X < 110) = ______________________________
P(100 < X < 145) = ______________________________
17.3 | 18.4 | 20.9 | 16.8 | 18.7 | 20.5 | 17.9 | 20.4 | 18.3 | 20.5 |
19.0 | 17.5 | 18.1 | 17.1 | 18.8 | 20.0 | 19.1 | 19.1 | 17.9 | 18.3 |
18.2 | 18.9 | 19.4 | 18.9 | 19.4 | 20.8 | 17.3 | 18.5 | 18.3 | 19.4 |
19.0 | 19.0 | 20.5 | 19.7 | 18.5 | 17.7 | 19.4 | 18.3 | 19.6 | 21.4 |
19.0 | 20.5 | 20.4 | 19.7 | 18.6 | 19.9 | 18.3 | 19.8 | 19.6 | 19.0 |
20.4 | 17.3 | 16.1 | 19.2 | 19.6 | 18.8 | 19.3 | 19.1 | 21.0 | 18.6 |
18.3 | 18.3 | 18.7 | 20.6 | 18.5 | 16.4 | 17.2 | 17.5 | 18.0 | 19.5 |
19.9 | 18.4 | 18.8 | 20.1 | 20.0 | 18.5 | 17.5 | 18.5 | 17.9 | 17.4 |
18.7 | 18.6 | 17.3 | 18.8 | 17.8 | 19.0 | 19.6 | 19.3 | 18.1 | 18.5 |
20.9 | 19.3 | 18.1 | 17.1 | 19.8 | 20.6 | 17.6 | 19.1 | 19.5 | 18.4 |
17.7 | 20.2 | 19.9 | 18.6 | 16.6 | 19.2 | 20.0 | 17.4 | 17.1 | 18.3 |
19.1 | 18.5 | 19.6 | 18.0 | 19.4 | 17.1 | 19.9 | 16.3 | 18.9 | 20.7 |
19.7 | 18.5 | 18.4 | 18.7 | 19.3 | 16.3 | 16.9 | 18.2 | 18.5 | 19.3 |
18.1 | 18.0 | 19.5 | 20.3 | 20.1 | 17.2 | 19.5 | 18.8 | 19.2 | 17.7 |
Mean : ___________________________.
Median : ___________________________.
Mode : ___________________________.
Standard Deviation : ___________________________.
P(Z < -1.33): _______________________
How would you interpret this probability? Discuss.
How would you interpret this probability? Discuss.
How would you interpret this probability? Discuss.
P(X > 117): _________________________(where X is the IQ score).
P(X < 72): _________________________ (where X is the IQ score).
P(85 < X < 115): _________________________ (where X is the IQ score).
50^{th} Percentile IQ score: _________________________ (where X is the IQ score).
85^{th} Percentile IQ score: _________________________ (where X is the IQ score).
SAT I Mean Scores and Standard Deviations for Males, Females, and Total by Ethnic Group
Ethnic Group | 82^{nd} Percentile Scores |
American Indian or Alaskan Native | |
Asian, Asian American, or Pacific Islander | |
African American or Black | |
Mexican or Mexican American | |
Puerto Rican | |
Latin American, South American, Central American, or Other Hispanic or Latino | |
White | |
Other |
Reference: http://lib.stat.cmu.edu/DASL/Datafiles/MilitiamenChests.html
Frequency, f | Chest Size (inches) |
3 | 33 |
18 | 34 |
81 | 35 |
185 | 36 |
420 | 37 |
749 | 38 |
1073 | 39 |
1079 | 40 |
934 | 41 |
658 | 42 |
370 | 43 |
92 | 44 |
50 | 45 |
21 | 46 |
4 | 47 |
1 | 48 |
Provide a hard copy of the graph.
Mean: _________________________.
Note: To enter the chest size values into MINITAB in a single column, we need to generate 3 values of 33, 18 values of 34, 81 values of 35, etc in separate columns and then stack them in one column. Use the sequence Calc® Make Patterned Data® Arbitrary Set of Numbers to generate the different sets of chest sizes. For instance, in order to generate 3 values of 33 in column C1, the following dialog box shows the appropriate entries.
Repeat for all other chest sizes and save in columns C2 to C16.
Next stack the generated values in column C17 by selecting Manip® Stack ® Stack Columns and fill in the appropriate boxes in the dialog box as shown below.
This stack all the observed chest size values in column C17. Now descriptive statistics can be computed for all the chest sizes in column C17.
Note: To compute the population standard deviation from the sample standard deviation given by the descriptive statistics in MINITAB, you need to multiply it by . For this data set n = 5738.
Standard deviation (s X): _____________________________.
Probability: _____________________________.
What assumption(s) are you making in computing the probability? Refer to parts (b) and (d). Discuss.
Probability: _____________________________.
Display the mean chest size and the value of the given chest size and shade the appropriate probability (area).
Probability: _____________________________.
Display the mean chest size and the values of the given chest size and shade the appropriate probability (area).
Mean, m | Standard deviation, s | P(m - s < X < m + s ) | P(m - 2s < X < m + 2s ) | P(m - 3s < X < m + 3s ) |
100 | 16 | |||
34 | 3.5 | |||
8.3 | 0.125 | |||
55 | 5 |
Approximately 68% of the observations will fall in the interval (m - 1s, m + 1s)
Approximately 95% of the observations will fall in the interval (m - 2s, m + 2s)
Approximately 99.7% of the observations will fall in the interval (m - 3s, m + 3s)
P(m - 1s < X < m + 1s) = ____________________________
P(m - 2s < X < m + 2s) = ____________________________
P(m - 3s < X < m + 3s) = ____________________________
Note: An exponential distribution is positively skewed for a random variable X ³ 0. The density function is given by
Next we need to generate exponential densities values for the generated values in column C1. To achieve this, select Calc® Probability Distributions® Exponential. Fill in the dialog box as shown below such that the densities are saved in column C2. Observe that we are using a mean of 1200.
Next plot the values in C2 versus the values in C1. Let the values in C2 be along the vertical (Y) axis and the values in C1 be along the horizontal (X) axis.
Provide a hard copy of the graph with this assignment.
Requested Percentages | Percentages |
% that are expected to fail before 200 hours | |
% that are expected to fail before 2000 hours | |
% that are expected to fail after 500 hours | |
% that are expected to fail after 1500 hours | |
% that are expected to fail between 500 and 1500 hours | |
% that are expected to fail between 1000 and 2000 hours |
Recall a binomial random variable X with parameters n and p has a
where n is the number of trials in the binomial experiment and p is the constant probability of success for each trial.
Suppose that a normal random variable X^{*} has the same mean and standard deviation as the binomial random variable.
MINITAB can be used to demonstrate that under certain minimal conditions, such as np > 5 and n(1 ? p) > 5, the normal and the binominal distributions have very similar shapes and hence would indicate that the corresponding probabilities are approximately equal. The approximation gets better with increase n.
Also, a probability of the form for the binomial distribution can be approximated by from the corresponding normal distribution.
Note: The 0.5 value that was added is called the continuity correction factor that is applied when converting from a discrete distribution to a continuous distribution.
A recent UCLA Center for Communication Policy found that 76% of all Internet users check their e-mails daily. Suppose that 200 Internet users are selected at random and let X represent the number of Internet users who check their e-mails daily.
Note: the mean m = 200´ 0.76 = 152 and the standard deviation
Let X^{*} be a normal random variable with the same mean and standard deviation as above.
Note: Recall for the normal distribution, virtually all (99.7%) of the data values will be within 3 standard deviations of the mean. Thus, virtually all the values will be between 152 - 3´ 6.0399 = 133.8803 and 152 + 3´ 6.0399 = 170.1197. For all practical purposes, almost all the normal values will lie between 133 and 171.