I. The inside diameter of a piston ring is normally distributed with the mean of
ID: 3361858 • Letter: I
Question
I. The inside diameter of a piston ring is normally distributed with the mean of 9.5 c standard deviation of 0.4 cm. em and a (a) Find the probability that the average diameter of 10 rings is in the interval (9.4,9.6), (b) Find the probability that the average diameter of 10 rings is in the interval (9.3,9.7), (c) Find the probability that the average diameter of 16 rings is in the interval (9.4,9.6) (d) Find the probability that the average diameter of 16 rings is in the interval (9.3,9,7). (e) The probability that the average diameter of 10 rings exceeds k is 0.9. Find k. ii (f) Find the smallest n so that the average diameter of n rings is in the interval (9.4,9.6) with probability at least 0.95Explanation / Answer
a) for n=10;
std error of mean =std deviation/(n)1/2 =0.4/(10)1/2 =0.1265
hence P(9.4<X<9.6)=P((9.4-9.5)/0.1265<Z<(9.6-9.5)/0.1265)=P(-0.7906<Z<0.7906)=0.7854-0.2146=0.5708
b)P(9.3<X<9.7)=P((9.7-9.5)/0.1265<Z<(9.3-9.5)/0.1265)=P(-1.5811<Z<1.5811)=0.9431-0.0569=0.8862
c)for n=16 ;
std error of mean =std deviation/(n)1/2 =0.4/(16)1/2 =0.1
P(9.4<X<9.6)=P((9.4-9.5)/0.1<Z<(9.6-9.5)/0.1)=P(-1<Z<1)=0.8413-0.1587 =0.6827
d)P(9.3<X<9.7)=P((9.7-9.5)/0.1<Z<(9.3-9.5)/0.1)=P(-2<Z<2)=0.9772-0.0228 =0.9545
e) here for 10th percentile ; zscore =-1.2816
therefore corresponding value =mean +z*Std deviaiton =9.5-1.2816*0.1265=9.3379
(ii) here margin of error E=(9.6-9.4)/2 =0.1
for 95% confidence interval ; z score =1.96
therefore required sample size n=(z*std deviation/E)2 =(1.96*0.4/0.1)2 =~ 62