Normal Random Variables and the Normal Approximation to the Binomial Chapter 4-3 Normal Random Variables and the Normal Approximation to the Binomial

Density curve In this chapter, we’ll learn to find probability of events with infinitely many outcomes. For instance, suppose we are interested in the length of telephones calls made by a sales staff. The length of calls could be anything. From 2 minutes to 40 minutes. Using data which has been collected for a large number of calls, we can determine a curve called a density curve. The basic property of a density curve is that it relates probability to area. If we want to know the probability that a random call lasts between 5 and 18 minutes, then we can look at the area under the density curve between 5 and 18. This is finding the probability of continuous random variables instead of discrete random variables from earlier chapters

Standard normal curve Properties of a Standard Normal Random Variable Let Z denote the standard normal random variable 1. The probability that Z takes a value between a and b, a<b, is the area under the standard normal curve between z=a and z=b. 2. Pr 𝑍≥0 =0.5 and Pr 𝑍≤0 =0.5 3. For each real number a, a>0, Pr⁡[−𝑎≤𝑍≤0=Pr⁡[0≤𝑍≤𝑎] 4. For each real number a, Pr 𝑍=𝑎 =0

Standard Normal Curve

Using the calculator Let Z be a standard normal random variable. Find the following probabilities: 1. Pr 𝑍 𝑖𝑠 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 𝑜𝑟 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 0.43 =Pr⁡[𝑍≤0.43] 2. Pr 𝑍 𝑖𝑠 𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 1.7 =Pr⁡[𝑍≥1.7] On the calculator, go to “Distribution” (2nd – vars), choose normalcdf. Input lower bound, upper bound, mean, and standard deviation. For negative infinity, use −1 𝑒 99 For positive infinity, use 1 𝑒 99

Example 1 Let Z be the standard normal random variable. Find Pr 𝑍 𝑖𝑠 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 −1.2 =Pr⁡[𝑍≤−1.2]

Example 2 Find Pr 𝑍 𝑖𝑠 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 0.2 𝑎𝑛𝑑 1.2 =Pr⁡[0.2≤𝑍≤1.2] Let Z be the standard normal random variable. Find Pr 𝑍 𝑖𝑠 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 0.2 𝑎𝑛𝑑 1.2 =Pr⁡[0.2≤𝑍≤1.2] Find Pr 𝑍 𝑖𝑠 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 −1.2 𝑎𝑛𝑑 0.2 =Pr⁡[−1.2≤𝑍≤0.2]

If we were to calculate all this by hand The Greek letter 𝜇 “Mu” is used to denote the mean. The Greek letter 𝜎 “Sigma” is used to denote standard deviation. The purpose of the work on the left, is it is used to convert data of a normal distribution back into the standard normal curve. 𝑎≤𝑋≤𝑏 𝑎−𝜇≤𝑋−𝜇≤𝑏−𝜇 𝑎−𝜇 𝜎 ≤ 𝑋−𝜇 𝜎 ≤ 𝑏−𝜇 𝜎

Example 3 A study of reaction times is conducted by repeating an experiment several times with each subject. Let a random variable be defined by assigning a time (in seconds) to each repetition of the experiment. Suppose X takes the values between 1.28 and 2.54, inclusive. Find the corresponding interval in which the random variable 𝑌=(𝑋 −1.7)/0.42 takes values. Also find the probability that X takes values between 1.28 and 2.54 This problem is telling us that 𝜇=1.7 and 𝜎=0.42

Example 4 Let X be a normal random variable with 𝜇=5 and 𝜎=10. Find the following probabilities: 1. Pr⁡[5≤𝑋≤15 2. Pr⁡[−10≤𝑋≤10

Example 5 The thermometers produced at Accutemp Corporation are tested for accuracy at 40℃. The testing is done on an automatic testing device which rejects any thermometers whose registered temperature differs from 40℃ by more than 1℃. After efforts at quality control are intensified, the rejection rate is 17.7 percent. That is, with probability 0.823 a randomly selected thermometer will show temperature between 39 and 41℃ when tested. Assume that the manufacturing process produces thermometers whose registered temperature at 40℃ is a normal random variable X with mean 40, and standard deviation 𝜎. 1. Find 𝜎 2. Find the probability that when a randomly selected thermometer is tested, it will show a temperature which differs from 40℃ by more than 2℃

Standard normal distribution

Example 6 An experiment consists of randomly selecting a male student at GSU and measuring his height. A random variable X is defined by associating with each student his height. Suppose that X is a normal random variable with mean 5 feet 10 inches, and standard deviation 2 inches. Find the percentage of male students with heights between 5 feet 6 inches and 6 feet 2 inches.

Normal approximation to a binomial random variable Recall that for a Bernoulli process: 𝜇=𝑛𝑝 𝜎= 𝑛𝑝(1−𝑝) With n = number of trials, and p = probability of success. Approximation method: For Pr⁡[𝑘≤𝑋≤𝑙] … Pr⁡[𝑘−0.5≤𝑌≤𝑙−0.5] To determine if this approximation method is valid: 𝑛𝑝>5 𝑎𝑛𝑑 𝑛(1−𝑝)>5

Example 7 A Bernoulli Process consists of 100 trials with success probability p=0.2. Find the probability that the binomial random variable takes a value between 25 and 29, inclusive.

Example 8 A survey indicates that 40% of the voters in Metroburg airport support a bond issue. At random, 150 different voters are sampled and asked questions about the bond issue. Assume the population of Metroburg is large enough that the sampling can be viewed as a Bernoulli process. Find the probability that at least 45 of those sampled support the bond issue.