A continuous random variable can take on any value in a specified interval or range


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1 Continuous Probability Distributions A continuous random variable can take on any value in a specified interval or range Example: Let X be a random variable indicating the blood level of serum triglycerides, measured in mg/dl The probability distribution of X is represented by a smooth curve called a probability density function 1
2 The total area under the probability density function (pdf) is equal to 1 If the pdf is represented by f(x), then f(x) dx = 1 The area under the curve between any two points x 1 and x 2 is the probability that X takes a value between x 1 and x 2 2
3 Instead of assigning probabilities to specific outcomes of the random variable X, probabilities are assigned to ranges of values The probability associated with any one particular value is equal to 0 Therefore, P(X = x) =0 Also, P(X x) =P(X>x) The probability density function indicates which ranges of values are more likely to occur than others 3
4 The cumulative distribution function of X is F(x) = P(X x) = x f(x)dx Its value is the area under the probability density function to the left of x The population mean and variance of a continuous random variable X have the same meaning as they did for a discrete (but not categorical) random variable They are used to summarize the behavior of the random variable in terms of a measure of location and a measure of dispersion 4
5 The expected value E(X), or µ, is the average value taken on by the random variable X E(X) = xf(x) dx Var(X), or σ 2, is the average squared distance of each possible value of X from µ Var(X) = (x µ)2 f(x) dx = E(X µ) 2 The standard deviation of X is σ = Var(X) 5
6 The Normal Distribution The normal distribution is the most widely used probability distribution in statistics It is also called the Gaussian distribution or the bellshaped curve Many random variables including blood pressure, weight, height, serum cholesterol level, and IQ score are approximately normally distributed However, the distribution s real importance will be seen in the areas of estimation and hypothesis testing The normal distribution can also be used as an approximation to many other distributions (such as the binomial) in various situations 6
7 The probability density function of a normal random variable X is given by f(x) = 1 2πσ e [ 1 2 ] (x µ) 2 σ 2 where <x< The symbol π (pi) represents a constant approximated by The letter e is also a constant, approximated by It is the base of the natural logarithms, such that ln(e x )=x 7
8 µ =E(X) σ 2 =Var(X) 8
9 µ and σ are the parameters of the normal distribution they completely define its shape 9
10 The normal distribution with mean µ and variance σ 2 is represented by N(µ, σ 2 ) To find P(X x), we would have to draw the probability density function of N(µ, σ 2 ) and determine the area to the left of x To do this, a table of areas calculated for the normal distribution can be used There is a problem here depending upon the application, the normal distribution could have any number of different values for µ and σ 2 We cannot tabulate every possible distribution In fact, only a single curve is tabulated 10
11 The standard normal distribution has mean 0 and variance 1, and is denoted by N(0, 1) Approximately 68% of the area under the standard normal curve lies between 1 and +1, about 95% between 2 and +2, and about 99% between 2.5 and
12 The cumulative distribution function for a standard normal curve is represented by where X N(0, 1) Φ(x) =F(x) =P(X x) 12
13 What is P( 1 X 1)? From column D, P( 1 X 1) = This is the probability that X takes a value within ±1 standard deviation of the mean 0 13
14 What is P(X 2)? From column B, P(X 2) = This is the probability that X takes a value at least 2 standard deviations above the mean Because the standard normal distribution is symmetric, P(X 2) = as well 14
15 For what value of x is it true that P( x X x) =0.95? From column D, x =1.96 The probability that X takes a value within ±1.96 standard deviations of the mean is 0.95 From column B, the probability that X takes a value more than 1.96 SD above the mean (or more than 1.96 SD below the mean) is
16 The 100 uth percentile of the standard normal distribution is represented by z u P(X <z u ) = u What is the 80th percentile of the standard normal distribution? We want the value z.80 for which P(X <z.80 ) = 0.80 From column A, P(X <0.84) = and P(X <0.85) =
17 What is the 90th percentile of the standard normal distribution? This time, we want the value z.90 for which P(X <z.90 ) = 0.90 From column A, P(X <1.28) = What is the 99th percentile? From column A, P(X <2.33) =
18 Suppose that X N(10, 4) If 10 is subtracted from X, the distribution is shifted or translated X 10 N(0, 4) If X 10 is then divided by 2, the spread of the distribution is changed (the distribution is rescaled) X 10 2 N(0, 1) In general, if X N(µ, σ 2 ) and Z = X µ σ then Z N(0, 1) 18
19 By transforming X into Z, the table of areas for the standard normal distribution can be used to estimate probabilities associated with X Note: E(X + c) = E(X)+c E(cX) = c E(X) Var(X + c) = Var(X) Var(cX) = c 2 Var(X) These are the same properties that applied to the sample mean x and sample variance s 2 19
20 Therefore, ( X µ E(Z) = E σ ) = 1 σ E(X µ) = 1 σ [E(X) µ] = 1 σ [µ µ] = 0 and ( X µ Var(Z) = Var σ ) = 1 Var(X µ) σ2 = 1 σ 2 Var(X) = 1 σ 2 (σ2 ) = 1 20
21 Example: The diastolic blood pressures of males years of age are normally distributed with µ = 80 mm Hg and σ 2 = 144 mm Hg 2 σ = 12 mm Hg Therefore, a diastolic blood pressure of = 92 mm Hg lies 1 standard deviation above the mean Individuals with blood pressures above 95 mm Hg are considered to be hypertensive What is the probability that a randomly selected male has a blood pressure above 95 mm Hg? ( ) X P(X >95) = P > = P(Z>1.25) =
22 Approximately 10.6% of this population would be classified as hypertensive The value z =1.25 is called a zscore A zscore quantifies how far the value of interest lies from the mean µ, measured in units of the standard deviation σ Note that = 15 mm Hg One standard deviation is 12 mm Hg Therefore, 15 mm Hg is or 1.25 standard deviations above the mean of 80 mm Hg 22
23 What is the probability that a randomly selected male has a diastolic blood pressure above 110 mm Hg? ( ) X P(X >110) = P > = P(Z>2.50) = Approximately 0.6% of the population has a diastolic blood pressure above 110 mm Hg The zscore of 2.50 indicates that 110 lies 2.50 standard deviations above the mean of 80 mm Hg 23
24 What is the probability that a randomly selected male has a diastolic blood pressure below 60 mm Hg? ( X 80 P(X <60) = P < 12 = P(Z< 1.67) = P(Z>1.67) = ) Approximately 4.8% of the population has a diastolic blood pressure below 60 mm Hg The zscore of 1.67 indicates that 60 lies 1.67 standard deviations below the mean of 80 mm Hg 24
25 What value of diastolic blood pressure cuts off the upper 5% of this population? Using column B of the table for the standard normal distribution, the value Z = cuts off an area of 0.05 in the upper tail We want the value of X that corresponds to Z =1.645 Z = X µ σ = X X = 99.7 Approximately 5% of the men in this population have a diastolic blood pressure greater than 99.7 mm Hg 25
26 What value of diastolic blood pressure cuts off the lower 10% of the population? Using column B, the value Z =1.28 cuts off an area of 0.10 in the upper tail of the standard normal distribution Therefore, Z = 1.28 cuts off an area of 0.10 in the lower tail We want the value of X that corresponds to Z = 1.28 Z = X µ σ 1.28 = X X = 64.6 Approximately 10% of the men in this population have a diastolic blood pressure lower than 64.6 mm Hg 26
27 In some situations the normal distribution can be used as an approximation to the binomial distribution The approximation is most accurate when the probability distribution of the binomial random variable X is fairly symmetric This happens when the number of independent trials n is fairly large and p is not too close to either 0 or 1 27
28 Since the binomial random variable X has mean np and variance npq, its distribution can be approximated by the normal distribution N(np, npq) Example: In the United States, the probability that an individual exercises regularly is 0.42 For a group of 50 persons, what is the probability that exactly 20 exercise regularly? The total number of individuals who exercise, X, has a binomial distribution with n =50 and p =0.42 P(X = 20) = ( 50) (0.42) 20 (0.58) = This is the exact binomial probability 28
29 Now note that np = 50(0.42) = 21 and npq = 50(0.42)(0.58) = 12.2 Let Y be a normal random variable with mean 21 and variance 12.2 P(X = 20) can be approximated by the area under the N(21, 12.2) curve that lies between 19.5 and 20.5 The 0.5 that is both added to and subtracted from 20 is called a continuity correction factor It helps to compensate for the fact that a discrete binomial distribution is being approximated by a continuous normal distribution 29
30 ) P(19.5 Y 20.5) = P( Z = P( 0.43 Z 0.14) = P(0.14 Z 0.43) = P(Z 0.43) P(Z 0.14) = = Similarly, the probability that exactly 7 individuals exercise regularly is approximated by the area under the normal curve that lies between 6.5 and 7.5 P(X = 0) is approximated by P(Y 0.5) P(X = 50) is approximated by P(Y 49.5) 30
31 When does the normal approximation to the binomial distribution work? In general, N(np, npq) may be used to approximate Bin(n, p) when npq 5 If a computer is available, no approximation is necessary 31
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