Adding 2 random variables that can be described by the normal model - PowerPoint PPT Presentation

phuong
adding 2 random variables that can be described by the normal model n.
Skip this Video
Loading SlideShow in 5 Seconds..
Adding 2 random variables that can be described by the normal model PowerPoint Presentation
Download Presentation
Adding 2 random variables that can be described by the normal model

play fullscreen
1 / 27
Download Presentation
Adding 2 random variables that can be described by the normal model
124 Views
Download Presentation

Adding 2 random variables that can be described by the normal model

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

  1. Adding 2 random variables that can be described by the normal model AP Statistics B

  2. Outline of lecture • Review of Ch 16, pp.376-78 (Adding or subtracting random variables that fit the normal curve)—last of Ch 16 • Follow along in the text • Remember that you can download this PowerPoint and a smaller, un-narrated one from the Garfield web site • Write down slide number if you don’t understand anything

  3. First thing: make a picture, make a picture, make a picture! • Well, actually, a table, but a table IS a kind of picture, right?

  4. Preliminaries: set-up and assumptions Setting up: NOTE: NOT time to get an entire box packed for sending; rather, we are only packing, not boxing! • P1=time for packing 1st stereo system • P2=time for packing NEXT stereo system • T=P1+P2 Assumptions: • Normal models for each RV • Both times independent of each other

  5. Step one: calculate expected value (aka find the mean) • Remember that the expected value (EV) is a fancy word for finding the mean. • And with the mean, the EV sum of two random variables is the sum of their means:

  6. Application to the packing and boxing problem • Mean (EV) of packing 1ST system is 9 min • Mean (EV) of packing 2nd system is also 9 min • Therefore:

  7. Calculate standard deviation just like we have before • Nothing new, same old formula:

  8. So what? • (You should always ask yourself “so what?” when somebody tells you to do something) • Well, we know that we had two RVs (random variables, not recreational vehicles; this is statistics, after all, not an auto show) that have a normal distribution. • So we now have a normal distribution and know the mean and standard deviation. • In statistical terms: N(18, 2.12) • Now we can evaluate this using what we learned in Ch. 6! A seriously big deal!

  9. Q: What is the probability that packing 2 consecutive systems takes over 20 min? • This is the question we need to answer. • Remember the z-formula from Ch 6? • Write it down, and we’ll apply it on the next slide.

  10. Setting up the problem • We already know the mean and standard deviation from our earlier calculations: 18 and 2.12, respectively. • The “y” value we are looking for is 20, so we set up the solution thusly:

  11. Are we done yet? • Of course not. • Our goal is to find the probability that it will take MORE than 20 minutes to package 2 consecutive stereo systems. • The z-value of 0.94 simply means that the area to the LEFT of that point on the z-table (text, A79-A80) will be the probability that packaging will take LESS than 20 minutes.

  12. Draw a picture! • The probability we get for z=0.94 is the dark blue area on the left. • However, we’re interested in the light blue area that’s ABOVE z=0.94

  13. But first….. • Remember that our table only measures the cumulative probability to the LEFT of the value. • That’s all we have, so let’s find it, and then answer the question more directly.

  14. How to use the table • It’s been a while, but get the X.x value on the inner column, and the 0.0x value across the top. The intersection is where the value lies, and looks like this:

  15. Finding the probability to the left of 0.94 • Since 0.94>0, look on p. A-80 • Find 0.90 along the z-column on the far left • Read across the top row to 0.04 • The intersection of the 0.90 row and the 0.04 is the percentage of the normal curve to the LEFT of 0.94….. • ……which should be 0.8294

  16. What are we looking for? • Not the blue area to the LEFT, but the clear area to the RIGHT of 0.94 • Calculate by subtracting the area to LEFT from 1: • 1.0000-0.8294=0.1706

  17. Why the difference from the textbook? • Beats me. • But 0.1706 isn’t all THAT different from 0.1736…3 parts in a thousand.

  18. Back to interpretation • We get a z-value of just over 17%. • That means, in everyday language, that there’s only a 17% chance (probability) that it will take more than 20 minutes to package 2 consecutive stereo systems. • NOTE: the AP exam expects you to write out things like this

  19. What percentage of stereo systems take longer to back than to box? Next question(bottom of page 377)

  20. Set-up on questions like this is crucial • The key is to realize that you don’t set it up as an inequality, exactly. • That is, the question is NOT P>B • Rather, the question is whether P-B>0. • We pick a different variable (D for “difference”) and define it as D=P-B

  21. Why do we do it like this? • We now can ask a specific question, namely what’s the expected value of D? • In statistical terms, we have E(D)=E(P-B). • We can now calculate these values using what we’ve learned in Ch 16 and combining it with the normal model from Ch 6.

  22. First the mean,then the standard deviation • E(D)=E(P-B)=E(P)-E(B) • E(P) we get from reading the mean for packing right off the table • We get E(B) the same way. • E(P)-E(B)=9 min – 3 min = 6 min

  23. Calculating the standard deviation • I like to calculate the SD directly, but you can start with the variance, and then take the square root. • Var(D)=Var(P-B)=Var(P)+Var(B) • =1.52+ 1.02 (from the table)=2.25+1=3.25 • σP-B=(3.25)½=1.8 min (approximately)

  24. Now we have the normal model • i.e., N(3, 1.80) • We are interested only in the values that are GREATER than 0, i.e., to the RIGHT of the value, like the purple:

  25. Now, to calculate • Same as we did before (Slide 9): • By table A-79, z=-1.67 has 0.0475 to its LEFT, but we want the area to the RIGHT • So subtract from 1 and get 0.9525.

  26. Interpreting the result • We have determined the percent of the time where P-B>0 • In other words, 95.25% of the time, it takes longer to pack the boxes than to box them. • How do we know? Because P-B is positive ONLY when P>B (otherwise, the difference would be negative)

  27. Exercise • Do Exercise 33 on p. 383-84 of the textbook • Take 10-15 minutes to complete all parts. • Review your answers with Ms. Thien, who has them all worked out.