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# LIFE SCI CALCULUS MATH 145

UW

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This 4 page Study Guide was uploaded by Addison Beer on Wednesday September 9, 2015. The Study Guide belongs to MATH 145 at University of Washington taught by Staff in Fall. Since its upload, it has received 70 views. For similar materials see /class/192108/math-145-university-of-washington in Mathematics (M) at University of Washington.

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Date Created: 09/09/15

MATH 145 Winter 2006 Normal Probability Density Functions 1 The Standard Normal Density function is fz 6722 27139 This is a probability density function with mean u 0 and standard deviation 039 1 The corresponding distribution function is important enough to be given a special notation z 0 wen Thus 2 is an antiderivative of The function 2 cannot be expressed in terms of the usual elementary functions Some calculators have a button to evaluate Tables of values for 2 can be found in most statistics books Many texts provide a table which lists the values of z for 0 S 2 S 3 Some tables give values of the function Nz z 7 05 NOTE ltIgt7a 1 7 ltIgta By the symmetry of the graph Thus if X is a continuous random variable whose probability distribution function is a normal curve with mean 0 and standard deviation 1 then Pa lt X lt b 2 dz ltIgtb i a 1 b 7 e x 27139 a Example 1 The area under the standard normal curve between the limits z 71 and z 3 is equal to ltIgt3 7 ltIgt71 617917 4601 15774 The table gives the value of ltIgt3 also ltIgt71 1 7 ltIgt1 The function 2 can be used to nd areas under a normal curve with mean u and standard deviation 039 This means nding the area under the curve l 1 my 5 2 0 039 27139 A change of variables letting s z 7 uU shows that 12 e2mdz 1 b 1 11 2 1 LEE 7 575 o dz 7 Ix27139 a x27T Lgli thus 1 b 1 2 b 7 545 d2 J i U 039 27139 a 039 039 If a quantity Z has a normal probability density function With mean u and standard deviation 0 then Pa lt Z lt b if i U Together with tables for values of ltIgtz or Nz7 this can be used to estimate what fraction of a normally distributed quantity lies in a given range Example 2 The heights of corn plants in a certain eld t a normal curve7 with mean 21 meters7 and standard deviation 55 meters Find the fraction of corn plants whose heights are in the range 16 to 20 meters Solution 2 In this problem7 the quantity Z is probability density function which is a normal curve with M 21 and 039 55 We need to nd P16 lt Z lt 20 20721 16721 147 147 P16 lt Z lt 20 55 55 lt1gt718 i 491 4286 i 1814 2472 Example 3 The body masses M of a population of sh in a lake are normally distributed about a mean of 225 grams with standard deviation 32 grams What fraction of the sh have masses in the range 200 to 250 grams Solution By the same steps as the previous example7 250 i 225 200 i 225 P200 lt M lt 250 T i lt1gt 32 lt1gt78 i lt1gt778 7823 i 2177 5646 Normal Approximations to Binomial Probabilities Binomial probabilities can be modelled after ipping a biased coin a certain number of times Precisely7 suppose that we have a biased coin meaning that the probabilities of heads may not be the same as the probability of tails7 and toss the coin n times We use the following notation p li robH7 q 1 7 p ProbT n number of coin tosses The probability of getting exactly k heads and thus 71 7 k tails is given by n k 7 nik Cnkp 1 p 7 where Cn7k M017 k For example7 if the coin has PH 727 and thus PT 287 and we toss the coin 25 times7 then the probability of getting exactly 20 heads equals C2520 7220 285 m 128 If this were plotted it would look very much like a normal curve We can nd the formula for the normal curve that approximates the above points using the following formulas for the mean and standard deviation for a binomial experiment7 117107 0 71190719 For the above coin toss p 72 and n 25 and thus M 2572 18 a 257228 224 We plot below the normal curve with M 18 and 039 224 that is the function 7 7z272f2 2 7 224 6 and we see that the curves are quite close Thus the area under the normal curve approximates the probability values for the bi nomial experiment The approximation is based on the following idea the probability of getting exactly k heads is approximately equal to the area under the normal curve between the values z k 7 i and z k This means z varies over an interval of length 1 centered at In the example of the coin toss experiment to approximate the probability of getting exactly 20 heads we would take the normal curve with M 18 and 039 224 and nd the area under this curve between z 195 and z 205 By the previous section this is 195 718 23924 lt1gt117 i lt1gt67 205 i 18 P195 lt X lt 205 W i lt1gtlt 8790 i 7486 1304 This is pretty close to the correct answer of 128 We have used the histogram correction The normal approximation to binomial probabilities becomes even more time saving in calculating the probability that the number of heads lies in some range For example the probability that the number of heads in the above experiment is at least 15 and at most 20 is approximated by the area under the curve between 145 and 205 using the histogram correction This calculation is 205 i 18 145 i 18 P145 lt X lt 205 7 W i W 08790 7 00594 8196 lt1gt117 i lt1gt7156 To calculate this using the binomial formula we would have to calculate C251572152810 125167216289 1251772l7288 O25187218287 125197219286 C25 207220285 This works out to be 0806 which is pretty close to the true answer The normal approximation becomes even better as 71 gets bigger however it is important that the mean M be not too close to the endpoints ln statistics the rule of thumb is The normal approximation to binomial experiments can be used provided that 711025 and 711710 25 This means that the mean 1 is between 5 and n 7 5 which forces 71 to be at least 10 Example 4 In the human population the probability of a live birth producing a male child is 52 and the probability that it produces a female child is 48 Use the normal approximation to answer the following If 100 children are randomly chosen what is the probability that the number of female children in the collection is at least 45 and at most 55 Solution 4 This is a binomial experiment since the experiment can be modelled by recording the gender of 100 children by ipping a biased coin 100 times In this case let p PF 48 and 1 7p PM 52 Then the mean and standard deviation are given by u np 10048 48 a i1004852 m 500 Note that the mean is between 5 and 95 so the rule of thumb says we can apply the normal approximation The normal approximation says that the probability of at least 45 females and at most 55 females is approximately equal to 555 7 48 445 7 48 14 7 14 5 505 7 54 9332 7 2420 6912 Example 5 Suppose that you toss a fair coin 1000 times Use the normal approximation to estimate the probability that the number of heads is between 490 and 510 Solution 5 This is binomial with n 1000 p 1 7p 5 Thus 0 10005 500 039 1100055 m 158 The normal approximation says that the probability that the number of heads is at least 490 and at most 510 is approximately 5105 7 500 4895 7 500 lt gt lt gt 158 158 lt1gt66 7 lt1gt766 7454 7 2546 4908

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