Upcoming SlideShare
×

# Applied Math 40S Slides Mar 22, 2007

1,511
-1

Published on

Working backwards from normal curve probabilities to z-scores, binomial distributions and binomial approximations to the normal curve.

0 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

Views
Total Views
1,511
On Slideshare
0
From Embeds
0
Number of Embeds
12
Actions
Shares
0
0
0
Likes
0
Embeds 0
No embeds

No notes for slide

### Applied Math 40S Slides Mar 22, 2007

1. 1. Case 3(a): Find the Z-Score that Corresponds to a Given Probability If we know the probability of an event, we can ﬁnd the z-score that corresponds to this probability. This is the reverse of what we did in Case 2. Sample question: What is the z-score if the probability of getting more than this z-score is 0.350? inuHorm(B.35) _ '. 385323-4?26 1nuH 1 rw| *~| or*m( . 65 ‘W26 Normal. Bell-shaped Curve
2. 2. Case 3(b): Find the Z-Value that Corresponds to a Given Probability What is the z-score if the probability of getting less than this z-score is 0.750? Normal. Be/ I-shaped Curve «to -30 -20 - +4_o Il. l‘IUl"<lD| ""FI': B.?5:' 11> _ . e.?4439?495 § - 0- 6775
3. 3. A multiple choice test has 30 questions. There are 4 choices for each question only one is correct. A student answers each question by randomly guessing. What is the probability of a student getting 12 questions right? binomPdF(3B.1/4, ' 7W3« (:3: «Wk 12) _B29%46392 L’. .. .,, ../ .6‘ ﬂtnum, r )’ «. ... -m What is the probability of a student getting at most 12 questions right? binompdf-(3B’1/4) L1 L2. L3 2 b1HONPd‘F(3@: :1. 735329992: -4.. . -W9 """ " {1- ?3582B9EI2I-: ‘4.. . \$0353 [.3535 HHS-‘)1. 2 man: {1.?8582B9B2E'4m ﬁg‘: Sum“ 978406359 Lam-=1 . 335323931.. . What is the probability of a student getting at least 12 questions right? sum(L - 3 , I . B5665828B1 Hns+Lz £1.73532B9B2E'4m I’IVI. EMJlI| -‘Q
4. 4. A multiple choice test has 30 questions. There are 4 choices for each question only one is correct. A student answers each question by randomly guessing. Determine all the probabilities for all possible outcomes and graph them as a histogram. P1=L1.-L2 Hotz P‘Iot3 laJIND0lIJ UP? Xm1n= @ ~: II= e= K‘. '. I-'! '! ><ma><=31 _ E 9 Id . ‘><sc_.1=1 X11st= L1 Vm1n= .B5 Freq= L2 Ynax= .2 Vsc1=.1 XPes=1 min=15 ri‘ndJc{2III n= E.353SE
5. 5. To Calculate the Mean and Standard Deviation of a Binomial Distribution n: #1xmz< “: “°P a= in-p-(1-p)) f= P(f¢1rea': :) -. :_ .5 J O, :- J_ ‘I; /, _/). _,= /,, ,,“m, / ‘ 0 9) -? ’’’/ i')(svS '-‘ ;7. g '-‘ 1?. 3 '7"/3-"’~ 3B*.25 ? .5 «l'(3B>I¢.2'5=I<. ?5) I S .7 A/ oR’*v7 c 2.3?1?as245 .7 ; >7>7:o)< If‘? my r7»°P2o it/ riff/7r-75. '1 [/ —}>)>/ § /2" / s / F 410/5 / ;3=3o(3,4.) "'(/ T/")=30(%§~) :7-_g‘ _ ‘ : '22_S
6. 6. F‘ 1=L1.-L2 r-'Iin=1Ii r-‘I-1:-c-:1.‘-' n= E.IJ33E '
7. 7. Using the Normal Approximation of a Binomial Distribution & Fnvb 2- \$‘<oRE' F'oR ””~ ‘°R7"€‘T 2 '-1% 33*. 25 7 5 . MINDUM -' l1—'2s r<3aa= .25»= .75> xm1n= —-5 2. Xmax= S 2.37/3 (12-7.5)/ Fins x5_. ;1=1 = l.37;¢, 1.89?366596 cumin: -_1 Vmax= l4 Vsc1=.1 Xres=1 FIr-ta= .?111 low: '5 I UP-=1.B5i-'3? This is very close to what we‘/ band using the binomial probabilities 3 slides back (().9784).
8. 8. How to tell if a binomial distribution can be approximated by a normal distribution n‘p25 and n‘(l-(3)25 or n-p°(l-p)2l0
9. 9. Are the following distributions normal approximations of binomial distributions? How do you know? (a) 60 trials where the probability of success on each trial is 0.05 ")""‘4/41\$’) : '5 W4 (b) 60 trials where the probability of success on each trial is 0.20 n/ °= Coﬁaza): /D. /> / -,-)-_6a/ Mal (c) 600 trials where the probability of success on each trial is 0.05 9? E. (d) 80 trials where the probability of success on each trial is 0.99
10. 10. Determine the mean and standard deviation for each binomial distribution. Assume that each distribution is a reasonable approximation to a normal distribution. (a) 50 trials where the probability of success for each trial is 0.35 / Q = "/9 7 50 (0.35): /? _ 5 0-_—Vn/ of/ .') : .(/ So/ o.3S)(0.‘5; = l . (b) 44 trials where the probability of failure for each trial is 0.28 I 3 Ts 1 I -’-"1 =96’ l. ?2 £31.43 0-: Yr 7:) ‘P / '1 r / 3 / /«'ag. m3 = ,_, _z (c) The probability of the Espro I engine failing in less than 50 00 km is 0.08. In 1998_, 16 000 engines were produced. Find the mean and standard deviation for the engines that did not fail. = 0 / Q: 4,; 3/}Ia(o.1zN O’-"l/ "“"l”‘5l°“’3l : /‘/ ?.{o ‘: -S‘/73/6‘),