Question

A soft-drink machine is regulated to discharge an average of 7 ounce per cup. The amount of drink is considered normally distributed with a standard deviation of 0.5 ounce. What is the probability that a cup chosen at random will contain 7. C2 7,8 a) between 6.5,andR2 ounce? b) exactly 7.0 ounce? c) less than 7.3 ounce?

Answer 1

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

(a) required answer is 0.8904

for x=6.2, z=(6.2-7)/0.5=-1.6

for x=7.8, z=(7.8-7)/0.5=1.6

P(6.2<X<7.8)=P(-1.6<Z<1.6)=P(Z<1.6)-P(Z<-1.6)=0.9452-0.0548=0.8904

P(Z<1.6)=0.9452( using ms-excel=normsdist(1.6))

P(Z<-1.6)=0.0548 ( using ms-excel=normsdist(-1.6))

(b) The normal distribution is a continuous distribution and hence the probability of getting exactly 7 is zero.

(c) answer is 0.7257

for x=7.3, z=(7.3-7)/0.5=0.6

P(X<7.3)=P(Z<0.6)=0.7257 ( using ms-excel=normsdist(0.6))

(d) answer is 0.0227

for x=8, z=(8-7)/0.5=2

P(cup will overflow)=P(X>8)=1-P(X<8)=1-P(Z<2)=1-0.9773=0.0227

(e) answer is 8.165

here we find x such that P(X>x)=0.01,

for this first we find z such that P(Z>z)=0.01

or, P(Z<z)=1-P(Z>z)=1-0.01=0.99

and z=2.33 and corresponding x=7+2.33*0.5=8.165

(g) answer is 0.9999

here z=(x^--mean)/(sd/sqrt(n))=(7.4-7)/(0.5/sqrt(25))=4

P(X^-<7.4)=P(Z<4)=0.9999