Question

The data shown represents the high scores for members of a bowling league over a particular season. Find the approximate percentile rank of a score of 267 by constructing a percentile graph. Score Frequency 249.5-254.5 5 254.5-259.5 16 259.5-264.5 22 264.5-269.5 12 269.5-274.5 5 A) 78 B) 72 C) 67 D) 82

Answer 1

right choice is D) 82

Class Interval	class mark (x)	frequency(f)	f*x	f*x2
249.5-254.5	252	5	1260	317520
254.5-259.5	257	16	4112	1056784
259.5-264.5	262	22	5764	1510168
264.5-269.5	267	12	3204	855468
269.5-274.5	272	5	1360	369920
	sum=	60	15700	4109860

mean=E(x)=sum(fx)/sum(f)=15700/60= 261.67

E(x²)=sum(fx²)/sum(f)=4109860/60=68497.67

var(x)=E(x²)=E(x)*E(x)=68497.67-261.67*261.67=28.22

sd(x)=sqrt(var(x))=sqrt(28.22)=5.31

now we use standard normal variate z=(x-mean)/sd

for x=267, z=(267-261.67)/5.31=1.00

P(Z<1)=0.8413

required percentile =100*P(X<267)=P(Z<1)=100*0.8431=84.31

nearest choice is D). 82

the percentile graph is given as