Lecture 14 (6.8 and 7.1) (1)

Math 1271: Section 6.8 Probability Page 5 of 7 Math12T] Lectwre 14 5. Median Another measure of a probability density function is the me- dian. Tfiefinitiorg The median m of a probability density function, is a number, defined as 1 o 1 . /f'i f(z)dz = 5 and /@ flz)dz = 5. Actually, median is a value separating the higher half from the lower half. Example 3: Find the median of the exponential distribution with mean p = 10. : "sdth, e f('l'): io t<o | -t "}'{8 " t2o - (0 t<co {_l_lo_e'lo {';0 m = =) fwde AN frrdt =0 % & - 00 s (=0°, m) f(f):o Thew m >0 g : ¢ - 4 :'f ('f,df -!-f H.)d{ i te [ fwd e [y | -t o s y lm ~ 1o | -to -7 o =-€ = (~1) {0 —'%:u[— . = [-e® e lo' - _2'_. w = Il m o ed h(e®) = In(% ) = Ine) = In(z) 2073'"(2) . il Scanned with CamScanner

Math 1271: Section 6.8 Probability Page 6 of 7 6. Normal Distribution ['Normal distribution [is defined as 1 _2;9)3 b f(z) = =y as X <b) =J ix) ol T PlasX K where u is the mean, measures the centre. o is the standard deviation, measuresthe spread. Jdeviation Stand ard Nm( D::cfihfmfrh fix) = e ar .' e '/T\ Example 4: IQ scores are distributed normally with meanyf 2 100 and standard deviation o = 15. What percentage of popula- tion has an IQ score between 85 and 1157 P(ss<X<lis) 7 = P12 Z<i)=2 Plo<2s)=2- 0,413 A . P(fls Z<t) yl\ R 0 L] ¢ C v I Mnm- Aot Examp?es%' Fmgl exam scores are determin€d to be norm F Aail distributed with mean p = 62 and standard deviation o = 6 'What percentage of students will achieve a score between 77 and 797 P(T"X""q )= P(25< 2'2'5) (onvert o {faadnd Mm{ab':hbm =Pos2428) -podndgers E Z= v}' = 0.4901- 04438 X=1 Z-= 11_;;12_--2;'; X=1 Z- 11';" = 2,83 W g G by Th B0 J G 307 " Cont. Scanned with CamScanner

Math 1271: Section 6.8 Probability Page 7 of 7 Practice 1: The weight W in grams of boxes of chocolate at the store are normally dl_strxbuted] with mean 200 grams and the standard deviation is 20 grams. What is the probability that the box of chocolate you buy weights more than 221 grams? 22l — 200 P(X>ZZI):O.6— Pl2ove X < 221) 2= =05 // \ =¢5 ~P(o< 2=105) 0.5~ g35a| 0- 1469 ) R § ~~ (D) Practice 2: Daily commuting time is normally distributed with mean 38.8 minutes and the standard deviation of 11.4 minutes. (a) What proportion of commuters have a daily commute of more than 1 hour? whyert o 2 -seve [ =bo = 2= fo=308 wl © P(T>b0) = PCE >4 = X =05~ P(o<2<18¢) ' = (;}-5—- 0.469 6 W i (b) What proportion of commuters have a daily commute be- tween 20 and 30 minutes? T -35.8 e [0 2= & = -1.¢5 Plao< T <30) ~p(-t65¢ F<w7t) e A | =30 Z - 20~ 39§ e = —off we den ke heflafiw dato — L /// 5 6% -o7] | 011 16d we have to hs'fiec{ itte Po(({'rw re«j«'au ;o»ny fgmmefiy, Pl-165< B < -011) = Plemt = 2 < )45) | & - cPlc<2Z <145) — Plo<2 <0-77) = 0.4505 — .279¢ = b 13 The End. Scanned with CamScanner