Test 1 - test01_s

# T he r ul s oft s pl yi ded a s pl m ean of 3 and a s

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Unformatted text preview: ul s oft s pl yi ded a s pl m ean of¯ = 3. and a s pl he es ent l es t he am e el am e x 12 am e 324.C om put t coe ci e he entofvari i oft s pl ecal hat t coe ci of at on he am e;r lt he ent vari ance ofs2 = 3. var at on ( V ) i deﬁned t be ii C s o St andar D evi i d at on × 100% M ean For t s s pl hi am e CV = CV 58% Sol i ut on: Fr t gi om he ven i or at on s = nf m i 1. 82 CV × 100% 58% 3. 12 3. 324 1. 82 STA 2028/T 1 ( pts) 4 – Page 4 of4 – N am e: 9.H ei sofadul ght twom en havea m ean of63. n.and a s andar devi i of2. n.W hatdoes Chebyshev's 6i t d at on 5i Theorem s about t per ay he cent age ofwom en w i hi 2 s andar devi i oft m ean? tn t d at on he A nsw er: A t l t eas 75 % oft hei s s he ght houl f lw i hi 2 s andar devi i oft m ean. d al t n t d at on he Sol i ut on: B y C hebys hev,t per he cent age f l s t f m ul or k = 2, olow he or a,f 1 13 1 1− 2 = 1− 2 = 1− = k k 44 A s a per cent age,t s i 75% hi s ( pts) 10.M ake a good dr i oft f equency hi t am oft f l i dat 10 aw ng he r s ogr he olow ng a. Cl s as Fr equency f 50–52 3 53–55 56–58 8 12 59–61 62–64 13 11 Instructi ons. D r t hi t am on t axes s t aw he s ogr he ys em bel , l ow abel t hor zont axi us ng t he i al s i he mi dpoi sofeach cl si er .Labelt ver i nt as nt val he t calaxi i ar st ckm ksas0,5,10,15.D r t hi t am so aw he s ogr t hatt e ar no gapsbet her e ween t ver i he t calbar .A tt t ofeach bar( ect s he op r angl ,pl t f equency e) ace he r ft or hat cl s as . Sol i ut on: W e cal at t m i cul e he dpoi s ofeach cl s and l nt as abelt hor zont he i alaxi us ng t s i hem . T he bar s ( r angl )ar cont guous and cent ed on t m i or ect es e i er he dpoi s nt . 15 12 10 13 11 8 5 3 0 48 51 54 57 60 63...
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