Data Analysis and Statistical Methods Statistics 651

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1 Data Analysis and Statistical Methods Statistics Suhasini Subba Rao Review of previous lecture We showed if S n were a binomial random variable, where n was large and the probability of a success and failure are not too close to zero, then the transformed (standardised) random variable Y n = S n mean of S n = Sn n π, standard deviation of S n n π (1 π) has a distribution which is close to a bell shaped curve which is centered about zero and has variance (or spread) equal to one. This is the standard normal distribution (though it is only true when n is large and π and (1 π) are not too small). 1 Review: evaluating probabilities from a normal distibution Suppose that Z N(0,1) (Z is a random variable with a standard normal distribution). This means that Z can take any number from negative infinite to positive infinite. However, most of the time you will draw between ( 3,3). Draw the normal distribution show what P( 3.0 Z 3) is on it. Evaluate this probability. We see that P( 3.0 Z 3) = P(Z 3) P(Z < 3) = , which is large. Hence we are likely to draw in this interval. Evaluate P(Z > 3.03). P(Z > 3.03) = This means with a very small probability we will draw greater than Given a probability, locating the value on the x-axis Suppose Z N(0, 1) (Z is a random variable with a standard normal distribution). Question 1 We want to find the value of t such that P(Z t) = 0.8. For example, if the weight of randomly selected person followed a standard normal distribution, then P(Z t) = 0.8, means the probability I draw a randomly selected person and their weight is less than t (we want to find this t) is equal to 0.8. Solution 1 To find the t, look inside the normal tables, and locate 0.8 (very, very important you look inside the table not on the sides). The read out and you will see that you should get the value Hence P(Z 0.85) = 0.8, and t = Question 2 We want to find the value of t such that P(Z > t) =

2 Solution 2 First draw this. We know that if P(Z > t) = 0.02, then P(Z t) = = 0.8. Look inside the tables to find 0.8. You should see approximately Hence P(Z 2.06) = 0.8. Hence t = Evaluating probabilities from other normal distribution For every mean µ and variance 2 we get a different normal distribution. See normal distribution introduction.pdf and guess the animal. Suppose that X is a random variable with, for example, mean 3 and variance 6. how do we evaluate P(X a)? One may think for every mean µ and variance 2 we require a new set of tables. This is not possible. In fact we only need the standard normal tables to evaluate any normal probability. In fact, like the binomial distribution with some shifting and squigging we can transform the non-standard normal into a standard normal distribution. 5 Standardisation Suppose a random variable has a normal distribution with mean µ and variance 2 we write this succinctly as X N(µ, 2 ). We transform it; Z = X µ, Then this new random variable Z N(0,1), is a standard normal distribution with mean zero and variance one Z N(0,1) (it is standard normal). We then have P(X < a) = P( X µ < a µ ) = P(Z < a µ ). Since a,µ and are known, a µ is just a number. Recall Z N(0, 1), hence to evaluate P(X < a) we need only to look it up in the standard normal tables. See the handout non standard normal.pdf. All the probabilities of a standard normal are known (use Table 1). 6 7

3 Some examples 1 Some examples 2 Suppose X N(,5): Suppose X N( 2,). Calculate (i) P(X >.), (ii) P(X 6), (iii) P(X < 3.2) and (iv) P(X 2.3). Calculate (i) P(3.2 < X 6), (ii) P(2.3 < X <.) and (iii) P(2.3 < X <. or 3.2 < X 6). For each probability draw the picture. Solutions with explanations and plots are given in examples normal probailities lecture.pdf. Calculate (i) P(X > 6), (ii) P(X 0), (iii) P(X < 1.2) and (iv) P(X < 2). Calculate (i) P( 6 < X 1.2), (ii) P(2 < X < 1.2). For each answer make a plot and indicate the probability of interest. A quick summary of the solutions is given below. 8 Solutions 2 Remember if X N( 2,), then the transformation Z = X+2 N(0,1) (standard normal). To each of the solutions below add your own picture. (a)(i) P(X > 6) = 1 P(X 6) = 1 P( X ) = 1 P(Z 3 ) = = (ii) P(X < 0) = P( X+2 2 ) = 0.2 (iii) P(X < 1.2) = P( X ) = (iv) P(X < 2) = P( X ) = (b)(i) P( 6 < X 1.2) = P(X < 1.2) P(X 6) = P( X ) P( X ) = (ii) This is a trick question! The event (2 < X < 1.2) can never arise (think about it). So P(2 < X < 1.2) = 0. Yet another example Suppose X N(10,16). Evaluate P(8 X 15). First normalise: subtract mean and divide by standard deviation (square root of variance): P(8 X 15) = P( 8 10 X ) ) = P( 1 2 Z 5 ) = P(Z 1.25) P(Z 0.5). Look up P(Z 1.25) and P(Z 0.5) in tables: P(Z 1.25) = 0.8 and P(Z 0.5) = Therefore P(8 X 15) = P( 1 2 Z 5 ) = P(Z 1.25) P(Z 0.5) = , which gives the desired answer

4 Finding P(X x) = 0.71 See x 5 = Solve this equation to give x = = 6.1. Question: Suppose X X(5, ), find the value of x such that P(X x) = Answer: In the same way we try to evaluate probabilities, we transform into a standard normal: Recalling that X X(5,) we have 0.71 = P(X x) = P( X 5 x 5 ) = P(Z x 5 ) = 0.71, where Z N(0,1). This means P(X 6.1) = You can check this! P(X 6.1) = P( X ) = P(Z 0.55), looking this up in the tables we get P(Z 0.55) = Hence it is correct. Look also at xaxis lecture.pdf, which gives a pictorial solutions for this example and the following two examples. Now look up P(Z y) = 0.71 in the tables. We found that y = This means P(X x) = P( X ) = x 5 ) = P(Z x 5 ) = P(Z Some examples 3 Question: Suppose X N(10,). Find the x such that P(X x) = Answer: We know from the density plot that P(X < x) = 1 P(X x) = = Hence it is easier, given the tables, to evaluate P(X < x) = 0.65, rather than P(X x) = P(X < x) = P( X 10 < x 10 ) = P(Z < x 10 ) = This gives, P(Z < y) = 0.65, looking up inside the tables we have y 0.3. Hence 0.3 = y = x 10. Solving this gives x = = P(X 11.17) = 0.35 Even more examples (a little bit harder) Question: Suppose Z N(0, 1). Find the x such that P( x Z x) = 0.5. Answer: We know that P( x Z x) = P(Z x) P(Z x). Now we know that P(Z x) = 1 P(Z > x). Now due to symmetry of the normal density we have P(Z > x) = P(Z x). Altogether this gives P( x Z x) = P(Z x) (1 P(Z x)) = 2P(Z x) 1 = 0.5. This means P(Z x) = 1.5/2 = 0.75 =

5 Looking up 0.75 in the table we have, P(Z 1.6) = 0.75, hence x = 1.6. Altogether this means P( 1.6 < Z < 1.6) = Question: (even harder - you do not have to solve this one). Suppose X N(5,). Find the x such that P(5 x X 5+x) =