Normal Distribution on the Calculator

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This page presents typical problems and the calculator solutions associated with finding probabilities and z-scores related to the normal distribution. In each case we will specify the mean and standard deviation of the distribution. This will be presented by the notation: N(mean,stddev) as in N(0,1) to represent "a normal distribution with mean=0 and standard deviation=1.

Typical problems:

For N(0,1), find P( x < 0.880).
For N(0,1), find P( x > 0.880).
For N(0,1), find P( ^–2.050 <x < 0.880).
For N(0,1), find P( x < ^–0.880 or x > 0.420).
For N(0.8,2.8), find P( x < 0.880).
For N(14.8,3.4), find P( x > 0.880).
For N(^–19.8,2.6), find P( ^–^–20.476 <x < ^–18.578).
For N(14.8,3.4), find P( x < 6.504 or x > 11.808).
Find the z-score that has an area of 0.9972 to its left.
Find the z-score that has an area of 0.9664 to its right.
Find the positive z-score such that there is a combined area of 0.246 to the left of the negative of the z-score and to the right of the positive z-score.
Find the positive z-score such that there is an area of 0.8262 between the negative of the z-score and the positive z-score.
Find the value y for a population that is N(4,2.5) such that P(x<y)=0.7517.

Figure 1	For N(0,1), find P( x < 0.880). We want the normalcdf( command. To get to it we use to open the DIST menu. Our command is in the second position of this menu. Highlight that choice and press .
Figure 2	We complete the command so that it reads normalcdf(-999,.88) and press . The result, .8105703864 is displayed. The command uses two arguements. The first gives the left or lower edge of the area that we want, the second gives the right or higher edge of the area that we want. We chose -999 just because it is so far to the left that we will, for all practical purposes, have the area of everything to the left of the value 0.88. When used with just two arguments, normalcdf( automatically assumes that the normal distribution has mean=0 and standard deviation=1. Also on Figure 2 we have a complete version of the normalcdf( command by adding the third and fourth parameters to make normalcdf(-999,.88,0,1). Of course, the result is the same.
Figure 3	While we are here we might as well investigate the effect of using -999 for the left side or our area. In Figure 3 we first rerun the command with the value -99 as the left edge of the area. There is no differnce in the answer. This really means that the area to the left of -99 is so small that it does not change the answer that we got, .8105703864. Then we ran the command again, this time using just -9 as the lower limit. Again, there is no change in the answer.
Figure 4	In Figure 4 we try another value, -5, but this time there is a change. THerefore, the area to the left of -5 is tiny but still large enough to change our other answers. When we try using -7 for the left edge we return to the original answer. Thus, as long as we use a left edge that is -7 or less we will not see a change in the computed answer on the calculator. Somewhat for emphasis, the rest of the examples used here will employ the -999 value as if it were negative infinity.
Figure 5	For N(0,1), find P( x > 0.880). For this and most of the subsequent problems, we will solve the problem two ways. First we will solve it in a fashion that corresponds with the approach we would take if we were looking up the answers on a table. After that we will solve it using the flexibility of the calcualtor. The most frequently supplied tables give the area to the left of the z-score. Here we want the area to the right of a specific value. Because the total area is 1 we get the area to the right by subtracting the area to the left from 1. Therefore, we form the command 1–normalcdf(-999,.88). From this we get the answer .1894296136. Of course, using the calcualtor command, we could use 999 as positive infinity and just write the command as normalcdf(,88,999). This produces the desired result.
Figure 6	For N(0,1), find P( ^–2.050 <x < 0.880). If we were doing this with a table we would find the entire area to the left of 0.88 and then subtract from that the area to the left of -2.050. This would give the area between the two
Figure 7	For N(0,1), find P( x < ^–0.880 or x > 0.420). Here we want the area that is not between the two values. That is we want the sum of the two tails. Following the system we would use with a table, we can get the left tail via normalcdf(-999,-0.88) and we can get the right tail via 1-normalcdf(-999,0.42). Then we just add the two values. With the calculator we can short-cut this via 1-normalcdf(-0.88,0.42).
Figure 8	For N(0.8,2.8), find P( x < 0.880). For this proble we have moved away from having a standard normal distribution. Now we have a normal distribution with mean=0.8 and standard deviation=2.8. For this we can use the 4-parameter version of the normalcdf( command. In this version we give the calculator the left edge, the right edge, the population mean, and then the population standard deviation. For this proble the desired command is normalcdf(-999,0.88,0.8,2.8), where we use the -999 to represent negative infinity.
Figure 9	Before we go to the next problem, it is worth noting that we could not have done the previous problem directly using the normal distribution table. The table gives values for a N(0,1) population and this problem has a N(0.8,2.8) population. To use a standard tble we would have needed to get a z-score from the problem. In this case that would be `z = (x - mu)/sigma = (0.88-0.8)/2.8 = 0.285714285714...`. Then we can look up that z-score on the standard normal distribution table. We could do the same on the calculator. First we compute (0.88-0.8/2.8) and then we get normalcdf(-999,Ans) where Ans holds the previous answer and is produced via the key sequence. This is shown in Figure 9. We note that we get exactly the same result.
Figure 10	For N(14.8,3.4), find P( x > 0.880). This is similar to the problem in Figure 5, but now with a N(14.8,3.4) population. To mimic an approach that would use a table, although we would have to compute the z-score to use a table, we use 1-normalcdf(-999,0.88,14.8,3.4). Or, we could use the more direct approach with normalcdf(0.88,999,14.8,3.4) to get the same result.
Figure 11	Figure 11 steps through the computations to exactly mimic using a table for the problem in FIgure 10. First we get the z-score. Then, from the table, we get the area to the left of that z-score. Finally, we get 1-Ans, resulting in the same value that we got more directly in Figure 10.
Figure 12	For N(^–19.8,2.6), find P( ^–^–20.476 <x < ^–18.578). This is just a slightly more complex version of the problem in FIgure 6. We solve it the same way but we need to use the 4-parameter version of the normalcdf( command. Taking the difference of the two "left side" commands requires a great deal of typing. Just writing the command takes up most of the calculator screen.
Figure 12a	Using the flexibility of the normalcdf( command to just ask for the area between the two values, as we have done in Figure 12a, takes much less typing.
Figure 13	For N(14.8,3.4), find P( x < 6.504 or x > 11.808). This is just a slightly more complex version of the problem in FIgure 7. We solve it the same way but we need to use the 4-parameter version of the normalcdf( command. Finding the sum of the two "tails" requires a great deal of typing. Just writing the command takes up most of the calculator screen.
Figure 13a	Using the flexibility of the normalcdf( command to just ask for 1 minus the area between the two values, as we have done in Figure 13a, takes much less typing.
Figure 14	Find the z-score that has an area of 0.9972 to its left. If we were using a table to do this we would be "reading the table backwards." That is, we would find the required "area", in this case 0.9972, in the table and then figure out which z-score corresponds to that table entry. On the calculatr, this is done via the invNorm( command. Just as with the normalcdf( command, the invNorm( has two forms. If we use it with one parameter then the invNorm( assumes that we are talking about a N(0,1) population and the command produces the appropriate z-score. Figure 19 talks about the 3-parameter version of the command. Figure 14 shows the DIST menu. We see that the invNorm( command is in position 3.
Figure 15	In Figure 15 we have formed the command invNorm(.9972) and the calculator responds with the value 2.770327233, meaning that if we were to ask for P(x<2.770327233) in a N(0,1) populations the answer would be 0.9972.
Figure 16	Find the z-score that has an area of 0.9664 to its right. We know that if the desired z-score has abn area of 0.9664 to its right, then it has 1-0.9664 to its left. Because the invNorm( command finds z-scores with the specified area to the left, we form the command invNorm(1-0.9664) and let the calculator do the work.
Figure 17	Find the positive z-score such that there is a combined area of 0.246 to the left of the negative of the z-score and to the right of the positive z-score. We know that the distribution is symmetric. The area to the left of the negative z-score will be the same as the area to the right of the positive z-score. We can solve this at least two ways. First, we can find the point on the right that has all but half of the 0.246 to its left. The command to do this is invNorm(1-0.246/2). Alternatively, we could find the opposite of the point on the left that has 0.9972 to ts 0.9972. The command to do this is -invNorm(0.246/2).
Figure 18	Find the positive z-score such that there is an area of 0.8262 between the negative of the z-score and the positive z-score. We know that the distribution is symmetric. The area to the left of the negative z-score will be the same as the area to the right of the positive z-score. We can solve this at least two ways. First, we can find the opposite of the left point that has half of the portion not in the 0.8262 to its left. The command to do this is -invNorm((1-0.8262)/2). Alternatively, we could find the right point that has half of all the area plus half of the desired area to its left. The command to do this is invNorm((.5+0.8262/2).
Figure 19	Find the value y for a population that is N(4,2.5) such that P(x<y)=0.7517. As noted above, the invNorm( command has an extended form where we can specify a mean and standard deviation for the popultion we are using. In such a case the invNorm( produces a value in the population that corresponds to the area under the curve to the left of that value. With values other than N(0,1), this is not a z-score because we always assume that the z-score is from a N(0,1) population. For the problem stated here the command invNorm(.7517,4,2.5) produces 5.699622803. Hterefore, in a N(4,2.5) distribution the value 5.699622803 has 0.7517 area under the curve to the left of that value.
Figure 20	We can get to this same answer in a different way. Starting with a N(0,1), what is the z-score for 0.7517? The command invNorm(0.7517) gives that answer as .6798491211. That really says that for a N(0,1) distributution, if we move to the point that is .6798491211 standard deviations above the mean, then we will have 0.7517 of the area to the left of that point. In the case of a distribution such as we have here, N(4,2.5), we need to move .6798491211 standard deviations above the mean. For us that means *Ans2.5+4**, which puts us at exactly the same point that we had in Figure 19.