Measurements that are normally distributed can be described in terms of the mean (or average value) µ and the standard deviation s. The measurements have the the following properties:

(1) The mean and mode both equal the median; that is, the average value and the most likely value are both in the middle of the distribution.

(2) The distribution of measurements is symmetric about the mean.

(3) Around 68.26% of the measurements should be within one standard deviation of average, around 95.44% should be within two standard deviations, and around 99.74% of the measurements should be within three standard deviations of average.

Such a random measurement X is called a normal random variable and is denoted by X ~ N(µ, s^2), where s^2 is the variance (or square of the standard deviation). When µ = 0 and s^2 = s = 1, then X is a standard normal random variable and is denoted by Z ~ N(0,1), rather than X.

The distribution of measurements follows a "Bell-Shaped" curve. Specifically, the graph of the probability density function (pdf) f(x) =
1/[s Sqrt(2Pi)] * e^[-(x-µ)^2 / (2 s^2)] is bell-shaped and is symmetric about the mean µ. The probability that measurements fall between values j and k is given by the area under this graph from point j to point k and is denoted by P(j <= X <= k).

The **NORMDIST** program can be used to compute probabilities such as P(X <=k), P(X >= k), or P(j <= X <= k). To execute the program, first enter **1**, **2**, or **3** to designate the type of probability you wish to compute. Next, enter the lower and upper bounds j and k (or just the single bound k for a tail probability). The program next asks if you want to see a shaded graph of the pdf. If so, enter **1**. If not, then enter **0**. After the graph, we receive a display of the desired probability.

** Example.** Heights of adults are generally found to be normally distributed. If the average height of women is 65.5 inches with a standard deviation of 2.5 inches, find the probability of adult women measuring from 64 to 68 inches tall.

*Solution.* After calling up the **NORMDIST** program, enter **3** for **MIDDLE PROB**, then enter **65.5** for MEAN, enter **2.5** for **STANDARD DEV.**, enter **64** for **LOWER BOUND**, and enter **68** for **UPPER BOUND**. Also enter **1** to see the graph.

We first receive a graph of the "Bell-Curve", centered at 65.5, with the region from 64 to 68 being shaded. Then we see that P(64 <= X <= 68) is about 0.5671. That is, around 56.71% of women should be from 64 to 68 inches tall.

**Other Features**:

1. If you entered **1** to see a graph, then the partially shaded graph of the Bell-Curve initially appears, but then is replaced by the display of the computed probability. To see the graph again, press **GRAPH**. Then to see the probability value again, press **CLEAR**. On the TI-86, press **CLEAR** twice, or just press **EXIT** to remove the graph. To exit the graph on the TI-89, press **HOME**. Then press **F5** to see the program output again.

2. If you initially enter either **1** or **2** to compute a tail probability, then both the left-tail and right-tail probabilities will be displayed in either case. However only the desired region will be shaded in the graph.

1. The heights of adult men are normally distributed with a mean of 69.5 inches and a variance of 7.025 inches. Find the probabilities that a man chosen at random will be (a) at least 72 inches tall, (b) at most 72 inches tall.

2. Scores on standard IQ Tests are usually designed to be normally distributed with a mean of 100 and a standard deviation of 15. On such a test, find the probability that a person chosen at random will score (a) below 90, (b) above 90.

3. On American Roulette wheels, the probability of the ball landing on red is 18 / 38. Suppose 200 bets are placed on red. Use the Normal Approximation of the Binomial to approximate the probability of there being from 100 to 120 winners.

4. It is estimated that Americans average 200 deaths yearly (per 100,000 people) from heart attacks. Use the Normal Approximation of the Poisson to approximate the probability that 180 to 210 such deaths will occur in a random group of 100,000 Americans during a given year.

1. Bring up the **NORMDIST** program, enter **2** for **RIGHT PROB**, then enter **69.5** for **MEAN** and enter **SQRT(7.025)** for **STANDARD DEV.** (use the square root since 7.025 is the variance). Next, enter **72** for **BOUND**. (If desired, enter **1** to see a graph. Otherwise enter **0**.) If graphing, we first receive a graph with a shaded right-tail region. Then we find that (a) P(X >= 72) = 0.1728 and (b) P(X <= 72) = 0.8272.

2. In this case, enter **1** for **LEFT PROB**, then enter **100** for **MEAN**, **15** for **STANDARD DEV.**, and **90** for **BOUND**. We find that (a) P(X < 90) = 0.2525 and (b) P(X > 90) = 0.7475.

3. For large n, we can approximate the binomial probability P(j <= Y <= k), where Y ~ b(n, p), with the following Normal probability: Let µ = np, s = Sqrt(np(1-p)), and let X ~ N(µ, s^2 ). Then P(j <= Y <= k) ~ P(j - .5 <= X <= k + .5).

Here, we have Y ~ b(200, 18 / 38). So, in the **NORMDIST** program, enter **3** for **MIDDLE PROB**, then enter **200*18 / 38** for **MEAN**, enter **Sqrt(200*18 / 38*20 / 38)** for **STANDARD DEV.**, enter **100 - .5** for **LOWER BOUND**, and enter **120 + .5** for **UPPER BOUND**.

We find that P(100 <= Y <= 120) ~ P(99.5 <= X <= 120.5) = 0.24985. That is, there is about a 25% chance of there being from 100 to 120 winners out of the 200.

4. We can approximate the Poisson probability P(j <= Y <= k), where Y ~ Poi(L), with the following Normal probability: Let µ = L, s = Sqrt(L), and let X ~ N(µ, s^2 ). Then P(j <= Y <= k) ~ P(j - .5 <= X <= k + .5).

Here, we can assume a Poisson Distribution Y with L = 200. So, in the **NORMDIST** program, enter **3** for **MIDDLE PROB**, then enter **200** for **MEAN**, enter **Sqrt(200)** for **STANDARD DEV.**, enter **180 - .5** for **LOWER BOUND**, and enter **210 + .5** for **UPPER BOUND**.

We find that P(180 <= Y <= 210) ~ P(179.5 <= X <= 210.5) = 0.6975. So there is nearly a 70% chance of there being from 180 to 200 such deaths in a group of 100,000 Americans during a given year.

**Note**: Exercises 3 and 4 may also be worked directly with the built-in **BINOMIAL** and **POISSON** programs.

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