Question

trd tsfolhs 200 Lal op accovlng ffatey af 3.4 an SE f2.2 ow Man f- thest Stndle t woldou Enpect to hou a ssA btwsn 3 5 andl Computal Aormal datrtion ith9ma a 1n the nears tenth

Answer 1

here n=200, mean=3.4, SE=2.2 ( I think this is standard deviation (SD) ans wrongly wrote SE)

(question is attempted on the assumption that it is SD, if not please respond back at earliest)

we use standard normal variate Z=(X-mean)/SD

for x=3.5, z=(3.5-3.4)/2.2=0.0455

for x=4.5, z=(4.5-3.4)/2.2=0.5

P(3.5<X<4.5)=P(0.0455<Z<0.5)=P(Z<0.5)-P(Z<0.0455)=0.6915-0.5181=0.1733

required number=n*P(3.5<X<4.5)=200*0.1733=34.66 ( next whole number is 35)

answer is 35