CHAPTER 8. Normal distribution Introduction. A synonym for normal is Gaussian. The first thing to do is show that this is a density.

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1 CHAPTER 8 Normal distribution 8.. Introduction A r.v. is a standard normal written N, )) if it has density e x /. A synonym for normal is Gaussian. The first thing to do is show that this is a density. Let I = e x / dx. Then Changing to polar coordinates, I = I = ˆ ˆ ˆ π/ ˆ e x / e y / dx dy. re r / dr = π/. So I = π/, hence e x / dx = as it should. Note ˆ xe x / dx = by symmetry, so E Z =. For the variance of Z, we use integration by parts: E Z = ˆ x e x / dx = ˆ x xe x / dx. The integral is equal to Therefore VarZ = E Z =. ] xe x / ˆ + e x / dx =. We say X is a N µ, ) if X = Z + µ, where Z is a N, ). We see that F X x) = PX x) = Pµ + Z x) = PZ x µ)/) = F Z x µ)/) if >. A similar calculation holds if <.) Then by the chain rule X has density f X x) = F Xx) = F Zx µ)/) = f Zx µ)/). This is equal to e x µ) /. 9

2 9 8. NORMAL DISTRIBUTION E X = µ + E Z and VarX = VarZ, so E X = µ, VarX =. If X is N µ, ) and Y = ax + b, then Y = aµ + Z) + b = aµ + b) + a)z, or Y is N aµ + b, a ). In particular, if X is N µ, ) and Z = X µ)/, then Z is N, ). The distribution function of a standard N, ) is often denoted Φx), so that Φx) = ˆ x e y / dy. Tables of Φx) are often given only for x >. One can use the symmetry of the density function to see that this follows from Φ x) = Φx); Φ x) = PZ x) = = ˆ x ˆ x e y / dy e y / dy = PZ x) = PZ < x) = Φx). Example 8.: Find P X 4) if X is N, 5). Answer. Write X = + 5Z. So P X 4) = P + 5Z 4) = P 5Z ) = P. Z.4) = PZ.4) PZ.) = Φ.4) Φ.) =.6554 [ Φ.)] =.6554 [.5793]. Example 8.: Find c such that P Z c) =.5. Answer. By symmetry we want c such that PZ c) =.5 or Φc) = PZ c) =.975. From the table we see c =.96. This is the origin of the idea that the 95% significance level is ± standard deviations from the mean. Proposition 8.: We have the following bound. For x > PZ x) = Φx) x e x /.

3 Proof. If y x, then y/x, and then 8.. INTRODUCTION 93 PZ x) = ˆ e y / dy x ˆ y / x e y dy x = x e x /. This is a good estimate when x 3.5. In particular, for x large, PZ x) = Φx) e x /.

4 94 8. NORMAL DISTRIBUTION 8.. Further examples and applications Example 8.3: Suppose X is normal with mean 6. If P X > 6) =.8, then what is the standard deviation of X? Answer. We apply the fact that X µ = Z is N, ) and get X 6 P X > 6) =.8 P > 6 6 ) =.8 P Z > ) =.8 P Z ) =.8 ) Φ =.8 ) Φ =.977. Using the standard normal table we see that Φ ) =.977, thus we must have that = and hence = 5. Copyright 7 Phanuel Mariano, Patricia Alonso-Ruiz.

5 8.3. EXERCISES Exercises Exercise 8.: Suppose X is a normally distributed random variable with µ = and = 36. Find a) P X > 5), b) P 4 < X < 6), c) P X < 8). Exercise 8.: The height of maple trees at age are estimated to be normally distributed with mean cm and variance 64 cm. What is the probability a maple tree at age grows more than cm? Exercise 8.3: The peak temperature T, in degrees Fahrenheit, on a July day in Antarctica is a Normal random variable with a variance of 5. With probability.5, the temperature T exceeds degrees. a) What is PT > 3), the probability the temperature is above freezing? b) What is P T < )? Exercise 8.4: The salaries of UConn professors is approximately normally distributed. Suppose you know that 33 percent of professors earn less than $8,. Also 33 percent earn more than $,. a) What is the probability that a UConn professor makes more than $,? b) What is the probability that a UConn professor makes between $7, and $8,? Exercise 8.5: Suppose X is a normal random variable with mean 5. If P X > ) =.8888, approximately what is V ar X)? Exercise 8.6: The shoe size of a UConn basketball player is normally distributed with mean inches and variance 4 inches. Ten percent of all UConn basketball players have a shoe size greater than c inches. Find the value of c. Exercise 8.7: The length of the forearm of a UConn football player is normally distributed with mean inches. If ten percent of the football team players have a forearm whose length is greater than.5 inches, find out the approximate standard deviation of the forearm length of a UConn football player. Exercise 8.8: Companies C and A earn each an annual profit that is normally distributed with the same positive mean µ. The standard deviation of C s annual profit is one third of its mean. In a certain year, the probability that A makes a loss i.e. a negative profit) is.8 times the probability that C does. Assuming that A s annual profit has a standard deviation of, compute approximately) the standard deviation of C s annual profit. Exercise 8.9: Let Z N, ), that is, a standard normal random variable. Find probability density for X = Z. Hint: first find the cumulative) distribution function F X x) = P X x) in terms of Φx) = F Z x). Then use the fact that the probability density function can be found by f X x) = F X x), and use the known density function for Z.

6 96 8. NORMAL DISTRIBUTION 8.4. Selected solutions Solution to Exercise 8.A): P X > 5) = P Z > 5 ) = P Z >.8333) 6 = P Z.8333) = Φ.8333) = Φ.8333) =.7977 Solution to Exercise 8.B):Φ) =.687 Solution to Exercise 8.C): Φ.3333) =.3695 Solution to Exercise 8.: We have µ = and = 64 = 8. Then ) P X > ) = P Z > = P Z >.5) 8 = Φ.5) =.56. Solution to Exercise 8.3A): We have = 5 = 5. Since P X > ) =.5 then we must have that µ = since the pdf of the normal distribution is symmetric. Then ) 3 PT > 3) = P Z > 5 = Φ.47) =.78. Solution to Exercise 8.3B): We have PT < ) = Φ.67) = Φ.67) =.54. Solution to Exercise 8.4A): First we need to figure out what µ and are. Note that ) 8, µ P X 8, ) =.33 P Z < =.33 ) 8, µ Φ =.33 and since Φ.44) =.67 then Φ.44) =.33. Then we must have Similarly, since 8, µ =.44. P X >, ) =.33 P X, ) =.33 ), µ Φ =.33 ), µ Φ =.67 Now again since Φ.44) =.67 then, µ =.44.

7 Solving the equations 8, µ =.44 and simultaneously we have that Then Solution to Exercise 8.4B): We have 8.4. SELECTED SOLUTIONS 97, µ µ =, and 45, P X >, ) =.5. P 7, < X < 8, ).753. =.44, Solution to Exercise 8.5: Since P X > ) =.8888, then P X > ) =.8888 P Z > 5 ) =.8888 P Z 5 ) =.8888 Φ 5 ) =.8888 )) 5 Φ =.8888 ) 5 Φ = Using the table we see that Φ.) =.8888, thus we must have that 5 =. and solving this gets us = 4.98, hence 6.8. Solution to Exercise 8.6: Note that P X > c) =. P Z > c ) =. P Z c ) =. P Z c ) =.9 ) c Φ =.9 Using the table we see that Φ.8) =.9, thus we must have that and solving this gets us c = c =.8

8 98 8. NORMAL DISTRIBUTION Solution to Exercise 8.7: Let X denote the forearm length of a UConn football player and let denote its standard deviation. From the problem we know that X PX >.5) = P >.5 ).5 ) = Φ =. From the table we get hence.479. Solution to Exercise 8.8: Let A and C denote the respective annual profits, and µ their mean. Form the problem ) we know PA < ) =.8PC < ) and A = µ/3. Since they are µ normal distributed, Φ =.8Φ 3) which implies µ Φ =. +.8Φ3).998. ) From the table we thus get µ/.88 and hence the standard deviation of C is µ/3 9.6.