Welcome Back Please hand in your homework. HW #5 Due next Thursday. Assignment is on the web. See website (on “notes” page) for some notes about how to use TI calculators, Excel, and Minitab to compute probabilities.

Determining normal probabilities: Suppose X has a normal distribution with mean 5 and std dev 2. Notation X~N(5,4) [notation uses N(mean,variance)] What’s the probability that X is less than 4?

Normal Density Pr(X<4) = area under curve to left of x=4 7

Manipulate to get to Pr(0<Z<a) and known quantities: What’s Pr(X < 4)? Draw (on next page) Center and scale: Pr(X<4) = Pr( (X-5)/2 < (4-5)/2 ) = Pr( Z < -1/2 ) Manipulate to get to Pr(0<Z<a) and known quantities: = Pr( Z < -1/2 ) = Pr(Z > 1/2) = 1-Pr(Z<1/2) = 1- [Pr(Z<0) + Pr(0<Z<1/2)] Look up = 1- [Pr(Z<0) + Pr(0<Z<1/2)] = 1-0.5-0.1915 = 0.3085

Pr(X<4) = area under curve to left of x=4 7

Example 2 First let’s do this with the tables. X ~ N(2,9) Want area in between these bars X ~ N(2,9) What’s Pr(1<X<4)? First let’s do this with the tables.

=Pr[(1-2)/3<Z<(4-2)/3] (where Z~N(0,1)) Pr(1<X<4) =Pr[(1-2)/3<Z<(4-2)/3] (where Z~N(0,1)) =Pr[Z<(4-2)/3] –Pr[Z<(1-2)/3] =Pr(Z < 2/3) – Pr(Z < -1/3) =Pr(Z<0) + Pr(0<Z<2/3) – Pr(Z>1/3) = Pr(Z<0) + Pr(0<Z<2/3) – (1-Pr(Z<1/3)) = Pr(Z<0) + Pr(0<Z<2/3) – [1-Pr(Z<0)-Pr(0<Z<1/3)] = 0.5 + 0.2486 – [1-0.5- 0.1293] = 0.3779 EVEN IF YOU’LL ALWAYS USE CALCULATORS, MINITAB, EXCEL, OR MATLAB TO DO THESE PROBLEMS, YOU’LL NEED TO KNOW HOW TO DO CALCULATIONS LIKE THE ONES ABOVE… Purpose of all this is to get to an expression that only uses Pr(0<Z<a) where a>0 and Z~N(0,1) and known quanities. All because tables have Pr(0<Z<a).

Using excel or minitab, the only step that is necessary is to get the probability in terms of CDFs (i.e. Pr[X <= k]). Pr(1<X<4) = Pr(X<4) – Pr(X<1), where X ~N(2,32) = 0.748 – 0.369 = 0.378 (Do demo in class) Three probabilities to memorize: Pr(Z < 2.33) = 99% Pr(Z < 1.96) = 97.5% Pr(Z < 1.28) = 90%

Let X~N(10,16). Find an a such that Pr(X < a) = 0.80. Plot of x versus Pr(X<x) when X~N(10,42) Later in the course, we will need to be able to do things like the following: Let X~N(10,16). Find an a such that Pr(X < a) = 0.80. Probabilities are on this axis a is this number here

Let X~N(10,16). Find an a such that Pr(X < a) = 0.80. =Pr(Z < (a-10)/4] =0.80 =Pr(0 < Z < (a-10)/4] = 0.30 Using the table “backwards” we find that Pr(0 < Z < 0.84) = 0.30 As a result, (a-10)/4 = 0.84 So, a = 13.36 This is called an inverse probability problem.

The Normal Distribution is Pervasive Examples of things that are normally distributed: Heights, weights, abilities, many, many other measurements In general, when a quantity is the result of a combination of many factors and influences, samples of that quantity are very likely to be approximately normally distributed. Why?

A store keeps track of the average amount spent by people each day. Let Xi = average amount spent on day i It turns out that there is a good reason to believe that Xi has a normal distribution!!! This reason is the CENTRAL LIMIT THEOREM

Central Limit Theorem Let X1,…,Xn be n independent random variables each with constant mean m and constant variance s2. Then, as n gets large, (X1+…+Xn)/n ~ N(m, s2/n) and (X1+…+Xn) ~ N(nm, ns2)

What does “large n” mean What “large” is depends on the distribution of Xi. If Xi’s are already normal, then the result is true for any n If Xi’s have a symmetric distribution, then n or 3 is probably large enough If Xi’s have a skewed distribution, the n of 20 or 30 is probably large enough

Example Suppose the amount of potassium in a banana is normally distributed with mean 630mg and standard deviation 40mg. You eat 3 bananas a day. Let T = amount of potassium you eat. What is the probability that T < 1800mg? By central limit theorem, T~N[3*630,3*(402)]. Want Pr(T < 1800) = Pr[ (T-1890)/(sqrt(3)*40) < (1800 – 1890)/(sqrt(3)*40)] = Pr[ Z < -1.33] = Pr[ Z > 1.33] = 1-Pr(Z < 1.33) = 1- Pr(Z < 0) – Pr(0 < Z < 1.33) = 1 – 0.5 – 0.4082 = 0.0981

Area under curve to left of line is Pr(T < 1800)