The Empirical Rule is just an approximation and only works for certain values. What if you want to find the probability for x values that are not integer multiples of the standard deviation? The probability is the area under the curve. To find areas under the curve, you need calculus. Before technology, you needed to convert every x value to a standardized number, called the z-score or z-value or simply just z. The
z-score is a measure of how many standard deviations an x value is from the mean. To convert from a normally distributed x value to a z-score, you use the following formula. Definition \(\PageIndex{1}\): z-score \[z=\dfrac{x-\mu}{\sigma} \label{z-score}\] where \(\mu\)= mean of the population of the x value and \(\sigma\)= standard deviation for the population of the x value The z-score is normally distributed,
with a mean of 0 and a standard deviation of 1. It is known as the standard normal curve. Once you have the z-score, you can look up the z-score in the standard normal distribution table. Definition \(\PageIndex{2}\): standard normal distribution The standard normal distribution, z, has a mean of \(\mu =0\) and a standard deviation of \(\sigma =1\). Luckily, these days technology can find probabilities for you without converting to the zscore and looking the probabilities up in a table. There are many programs available that will calculate the probability for a normal curve including Excel and the TI-83/84. There are also online sites available. The following examples show how to do the calculation on the TI-83/84 and with R. The command on the TI-83/84 is in the DISTR menu and is normalcdf(. You then type in the lower limit, upper limit, mean, standard deviation in that order and including the commas. The command on R to find the area to the left is pnorm(z-value or x-value, mean, standard deviation). Example \(\PageIndex{1}\) general normal distribution The length of a human pregnancy is normally distributed with a mean of 272 days with a standard deviation of 9 days (Bhat & Kushtagi, 2006).
Solution a. x = length of a human pregnancy b. First translate the statement into a mathematical statement. P (x>280) Now, draw a picture. Remember the center of this normal curve is 272. Figure for Example \(\PageIndex{1}\)bTo find the probability on the TI-83/84, looking at the picture you realize the lower limit is 280. The upper limit is infinity. The calculator doesn’t have infinity on it, so you need to put in a really big number. Some people like to put in 1000, but if you are working with numbers that are bigger than 1000, then you would have to remember to change the upper limit. The safest number to use is \(1 \times 10^{99}\), which you put in the calculator as 1E99 (where E is the EE button on the calculator). The command looks like: \(\text{normalcdf}(280,1 E 99,272,9)\) Figure \(\PageIndex{3}\): TI-83/84 Output for Example \(\PageIndex{1}\)bTo find the probability on R, R always gives the probability to the left of the value. The total area under the curve is 1, so if you want the area to the right, then you find the area to the left and subtract from 1. The command looks like: \(1-\text { pnom }(280,272,9)\) Thus, \(P(x>280) \approx 0.187\) Thus 18.7% of all pregnancies last more than 280 days. This is not unusual since the probability is greater than 5%. c. First translate the statement into a mathematical statement. P (x<250) Now, draw a picture. Remember the center of this normal curve is 272. Figure for Example \(\PageIndex{1}\)cTo find the probability on the TI-83/84, looking at the picture, though it is hard to see in this case, the lower limit is negative infinity. Again, the calculator doesn’t have this on it, put in a really small number, such as \(-1 \times 10^{99}=-1 E 99\) on the calculator. Figure \(\PageIndex{5}\): TI-83/84 Output for Example \(\PageIndex{1}\)c\(P(x<250)=\text { normalcdf }(-1 E 99,250,272,9)=0.0073\) To find the probability on R, R always gives the probability to the left of the value. Looking at the figure, you can see the area you want is to the left. The command looks like: \(P(x<250)=\text { pnorm }(250,272,9)=0.0073\) Thus 0.73% of all pregnancies last less than 250 days. This is unusual since the probability is less than 5%. d. First translate the statement into a mathematical statement. P (265<x<280) Now, draw a picture. Remember the center of this normal curve is 272. Figure for Example \(\PageIndex{1}\)dIn this case, the lower limit is 265 and the upper limit is 280. Using the calculator Figure \(\PageIndex{7}\): TI-83/84 Output for Example \(\PageIndex{1}\)d\(P(265<x<280)=\text { normalcdf }(265,280,272,9)=0.595\) To use R, you have to remember that R gives you the area to the left. So \(P(x<280)=\text { pnom }(280,272,9)\) is the area to the left of 280 and \(P(x<265)=\text { pnom }(265,272,9)\) is the area to the left of 265. So the area is between the two would be the bigger one minus the smaller one. So, \(P(265<x<280)=\text { pnorm }(280,272,9)-\text { pnorm }(265,272,9)=0.595\). Thus 59.5% of all pregnancies last between 265 and 280 days. e. This problem is asking you to find an x value from a probability. You want to find the x value that has 10% of the length of pregnancies to the left of it. On the TI-83/84, the command is in the DISTR menu and is called invNorm(. The invNorm( command needs the area to the left. In this case, that is the area you are given. For the command on the calculator, once you have invNorm( on the main screen you type in the probability to the left, mean, standard deviation, in that order with the commas. Figure \(\PageIndex{8}\): TI-83/84 Output for Example \(\PageIndex{1}\)eOn R, the command is qnorm(area to the left, mean, standard deviation). For this example that would be qnorm(0.1, 272, 9) Thus 10% of all pregnancies last less than approximately 260 days. f. From part (c) you found the probability that a pregnancy lasts less than 250 days is 0.73%. Since this is less than 5%, it is very unusual. You would think that either the woman had a premature baby, or that she may be wrong about when she actually became pregnant. Example \(\PageIndex{2}\) general normal distribution The mean mathematics SAT score in 2012 was 514 with a standard deviation of 117 ("Total group profile," 2012). Assume the mathematics SAT score is normally distributed.
Solution a. x = mathematics SAT score b. First translate the statement into a mathematical statement. P (x>700) Now, draw a picture. Remember the center of this normal curve is 514. Figure for Example \(\PageIndex{2}\)bOn TI-83/84: \(P(x>700)=\text { normalcdf }(700,1 E 99,514,117) \approx 0.056\) On R: \(P(x>700)=1-\text { pnorm }(700,514,117) \approx 0.056\) There is a 5.6% chance that a person scored above a 700 on the mathematics SAT test. This is not unusual. c. First translate the statement into a mathematical statement. P (x<400) Now, draw a picture. Remember the center of this normal curve is 514. Figure for Example \(\PageIndex{2}\)cOn TI-83/84: \(P(x<400)=\text { normalcdf }(-1 E 99,400,514,117) \approx 0.165\) On R: \(P(x<400)=\operatorname{pnorm}(400,514,117) \approx 0.165\) So, there is a 16.5% chance that a person scores less than a 400 on the mathematics part of the SAT. d. First translate the statement into a mathematical statement. P (500<x<650) Now, draw a picture. Remember the center of this normal curve is 514. Figure for Example \(\PageIndex{2}\)dOn TI-83/84: \(P(500<x<650)=\text { normalcdf }(500,650,514,117) \approx 0.425\) On R: \(P(500<x<650)=\text { pnorm }(650,514,117)-\text { pnorm }(500,514,117) \approx 0.425\) So, there is a 42.5% chance that a person has a mathematical SAT score between 500 and 650. e. This problem is asking you to find an x value from a probability. You want to find the x value that has 1% of the mathematics SAT scores to the right of it. Remember, the calculator and R always need the area to the left, you need to find the area to the left by 1 - 0.01 = 0.99. On TI-83/84: \(\text{invNorm}(.99,514,117) \approx 786\) On R: \(\text{qnorm}(.99,514,117) \approx 786\) So, 1% of all people who took the SAT scored over about 786 points on the mathematics SAT. HomeworkExercise \(\PageIndex{1}\)
1. a. \(P(z<2.36)=0.9909\), b. \(P(z>0.67)=0.2514\), c. \(P(0<z<2.11)=0.4826\), d. \(P(-2.78<z<1.97)=0.9729\) 3. a. -0.6667, b. -2.6667, c. -2, d. 6.6667 5. a. See solutions, b. \(P(x<52 \mathrm{cm})=0.7128\), c. \(P(x>74 \mathrm{cm})=5.852 \times 10^{-11}\), d. \(P(40.5 \mathrm{cm}<x<57.5 \mathrm{cm})=0.9729\), e. See solutions, f. 53.8 cm 7. a. See solutions, b. \(P(x>110 \mathrm{g})=0.0551\) c. \(P(x<93 \mathrm{g})=0.0097\), d. \(P(x<65 \mathrm{g}) \approx 0\) or \(5.57 \times 10^{-19}\), e. See solutions, f. 110.2 g 9. a. See solutions, b. \(P(x>\$ 80,000)=0.1168\), c. \(P(x>\$ 80,000)=0.2283\), d. \(P(\$ 55,000<x<\$ 72,000)=0.5519\), e. See solutions, f. $73,112 How do you find the probability with mean and standard deviation?Here x represents values of the random variable X, μ is the mean of X, P(x) represents the corresponding probability, and symbol ∑ represents the sum of all products (x−μ)2P(x). To find the standard deviation, σ, of a discrete random variable X, simply take the square root of the variance σ2.
How do you find the mean and standard deviation on a TI 84 Plus for probability distribution?Using the Calculator: To find the mean and standard deviation of a probability distribution, First: STAT > EDIT, then in L1 put in all the x values, and in L2 put in the probability for each x value. Second: STAT > CALC > 1-Var Stats > 1-Var Stats L1, L2 ENTER. 1.
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