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Chapter Five Normal Probability Distributions -Uniform Distribution: This means the probability desity function is a constant or zero. For example, throwing a die is a discrete uniform distribution where p(x = i ) = 1/6 for i = 1,2,3,4,5,6 and zero for any other value. You can also have a continuous uniform distribution. For example let p(x) = 1/2 for 2 ≤ x ≤ 4 and p(x) = 0 elsewhere. In both cases the sum of the probabilities is one. In the second case the ’area’ under the graph represents the sum. Question : Given that p(x) = A for 0 ≤ x ≤ 2 and for 5 ≤ x ≤ 6 and zero elsewhere, ﬁnd A. Solution: This is a continuous uniform distribution. The area underneath the graph MUST equal one. Thus 2A + A = 1 → A = 1 3 ––––––––––––––––––––––––––––– -Normal Distributions: This is a distribution which occurs often in nature, and also naturally occurs when we do averaging (see central limit theorem later). Given a particular µ and σ we get a particular Normal distribution. To ﬁnd the area underneath this curve (thus the probability) , we use the Standard Normal Distribution ( µ = 0, σ = 1). Every book on statistics has a chart for the area underneath this graph. For example: Find p(z > 0). (note : we use ’z’ instead of ’x’ when using standard normal - just helps keep things straight). To ﬁnd this value you can look it up in the standard normal chart or use common sense for this one....answer = 0.5 Here are a few more examples using standard Normal. ex) Find p(0 < z < 1.2) = 0.3849 ex) Find p(−1 < z < 2.39) = 0.3413 + 0.4916 = . 8329 ex) Find p(−2.24 < x < −1.75) = 0.4875 − 0.4599 = .0 276 –––––––––––––––––––––––––––– In real life we usually do not have mean of zero and standard deviation of one. But if we know the data is normally distributed we can transform the x−µ data using the formula z = σ . This allows use to use the standard normal distribution to ﬁnd probabilities. Example: Find p(x > 12.24) given that µ = 11 and σ = 3.2. 1 Solution: p(x > 12.24) = p(z > 12.24−11 ) = p(z > 0. 39) = 1 − 0.6517 = 0 3.2 . 3483 –––––––––––––––––––––––––––– More Examples: -Question: Lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. If we stipulate that a baby is premature if born at least three weeks early, what percentage of babies are born prematurely ? -Solution: p(x < 268 − 21) = p(x < 247) = p(z < 247−268 ) = p(z < 15 −1. 4) = 0.5 − 0.4192 = .0 808. So about 8.1% of babies are premature. -Question: An IBM subcontractor was hired to make ceramic substrates that are used to distribute power and signals to and from computer silicon chips. Speciﬁcations require resistance between 1.5 and 2.5 ohms, but the population has normally distributed resistances with a mean of 1.978 ohms and and a standard deviation of 0.172 ohms. What percentage of the ceramic substrates will not meet the namufactures’s speciﬁcations ? -Solution: p(x < 1.5) + p(x > 2.5) = p(z < 1.5−1.978 ) + p(z > 2.5−1.978 ) = 0.172 0.172 p(z < −2. 78) + p(z > 3.03) = (0.5 − 0.4973) + (0.5 − 0.4988) = .00 39. So about 0.39% of them will not meet the standards. ––––––––––––––––––––––––––– Going Backwards with the Normal distribution In these cases you are given a probability , say for example you wish to ﬁnd the mark you need to get into the top10% of your class. You would have to look at Standard Normal Chart ﬁrst and ﬁnd the corresponding ’z’ value, and equate this with z = x−µ . σ Example: The average mark in your graduating class is 65% with a stan- dard deviation of 10% .What mark would you require to make sure you get into the top 5% of your class. Solution: The corresponding z-value is 1.645, Thus x−65 = 1.645 → x − 10 65 = 16.45 → x = 81.45. So you would need about 82%. Example: Find P35 given that we have a mean of 43 and standard devi- ation of 6.2. Solution: The corresponding z-value is -0.385. Thus x−43 = −0.385 → 6.2 x − 43 = (6.2)(−0.385) = −2. 387 → x = 43 − 2.387 = 40. 613. Thus P35 = 40.61 ––––––––––––––––––––––––––– Central Limit Theorem 2 Deﬁnition: The Sampling Distribution of the mean is the probability distribution of sample means, with all samples having the sample size n. The Basic idea here is that as we take larger samples and take averages, the distribution starts to become Normally distributed. The problem is that the standard deviation of this sampling distribution changes and we must be cautious WHEN we can use this idea. Central limit Theorem Given - The random variable x has a distribution with mean µ and standard deviation σ (The distribution does not have to be normal). - Samples of size n are randomly selected from this population. (The samples are selected so that all possible samples of size n have the same chance of being selected). Conclusions: -The distribution of sample means x will, as the sample size increases, approach a normal distribution. -The mean of this distribution of sample means will be the population mean µ. (ie, µx = µ) σ -The standard deviation of this distribution of sample means will be √n . Notes: -For sample size n larger than 30, the distribution of the sample means can approximated reasonably well by a normal distribution (larger the n, the better the approximation. -If the original population is itself normally distributed, then the sample means will be normally distributed for any sample size n. σ -σ x = √n is called the standard error of the mean. –––––––––––––––––––––––— Examples of Central Limit Theorem: Question: Assume that female students’ heights are normally distributed witha a mean given by 64.2 inches and a standard deviation of 2.6 inches. If 4 female students are randomly selected, ﬁnd the probability that their mean height is between 63 and 65 inches. Solution: You can apply central limit theorem as the original distribution 2.6 is normally distributed. σ x = √4 = 1. 3. Thus we want p(63 < x < 65) = p( 63−64.2 < z < 65−64.2 ) = p(−0. 92 < z < 0. 62) = .3212 + .2324 = . 5536. So 1.3 1.3 probability of 55.36% 3 Question: Information provided by Bell CAnada showed that the average monthly bill for basic service was normally distributed with a mean of $15.30 and a standard deviation of $1.25. If 55 phone bills are randomly selected, ﬁnd the probability that their mean exceeds $15. 1.25 Solution: σ x = √55 = 0. 169. We want p(x > 15) = p(z > 15−15.3 ) = 0.169 p(z > −1. 78) = 0.9625. Therefore 96.25%. ––––––––––––––––––––––––– Correcting for a ﬁnite population: -It was assumed (in the ’given’ section for central limit theorem) that the population has inﬁnitely many members. This can be done by sampling with replacement - this eﬀectively makes the population inﬁnite. But, if we do not use replacement, and we have a ﬁnite population, our estimate for the standard deviation is slightly oﬀ. When sampling without replacement and the sample size n is greater than 5% of the ﬁnite population size N (that is n > 0.05 N), adjust the standard deviation of sample means σ x by multiplying N −n it by the ﬁnite population correction factor: N−1 . Question: As part of his review for a chemistry test, a student randomly selects 40 out of the 102 elements listed on the standard tables, and tries to guess their atomic numbers. Given that the mean of all 102 atomic numbers is 51.794, and the population standard deviation is 29.804, what is the prob- ability that the mean of the 40 atomic numbers selected by the student will be at least 58.3 ? Solution: σ x = 102−40 29.804 = 3. 69. We want p(x > 58.3) = p(z > 102−1 √ 40 58.3−51.794 3.69 ) = p(z > 1. 76) = 0.5 − 0.4608 = .0 392. Therefore a 3.92% chance. 4