ihTs qieuonts lnoy csrnceno nmowe ni eth 0-545 age or.pug eTh rguop sah a mean of 624 nad nddrasat doetainvi of .05 efeoTrr,eh lla otesh whit uesalv teraerg ahtn 926 are lla hseto bavoe neo rddsaant .eiitaodnv 2/3 of lal elavsu no a nlramo isibtndurtoi aer wihnti noe ddanarst vdatienio in ihtere rnteodcii. heero,fTer 3/1 aer edistuo fo shit ni tehrie ,ditoncire angeimn 13/ of newom hvae a evlau sesl hant 169 or tgerrae anht 962. If ew pilst atht ni fhal ot loyn osoceh hoste errtage tahn 2,96 ew egt 6/1 m,wneo hcihw is botua .6%1
loko at FA 1029 p.g 162
in a uaiaGnss, 66% rae niitwh 1 SD and 95% wiitnh 2 DS
hits is ihgh ydlie af
1 SD= 68% of the population 2 SD= 95% of the population 1 SD= 99.7% of the population
They tell us the standard deviation is +/- 50 from the mean.
So women b/w ages 50-54 had a mean of 246. 1 SD= 196-296 meaning 68% of females in this range are between this range.
That means 32% are not within this range (100-68=32). They are either above OR below this range.
The question is asking how many will be above 296. So you divide 32 by 2 so that you get the percentage of only those falling above 296.