Suppose that X = amount of soda in a 12oz-sized can of soda is a random variable

Question

Suppose that X = amount of soda in a 12oz-sized can of soda is a random variable which is normally distributed with mean mu = 12.2 and standard deviation sigma = 0.5. (a) Find the probability that a randomly chosen can of soda has between 11.7 and 12.0oz of soda. (b) Suppose the FDA decides that too many of the cans of soda have below 12oz of soda, and the beverage company must change their manufacturing practices so that at most 5% of cans have less than 12oz. Suppose that the beverage company keeps the mean soda mu = 12.2 but decreases the standard deviation sigma to conform to this restriction. How low must sigma be so that only 5% of cans have less than 12oz?

Explanation / Answer

We are given that P(X<12)=0.05 and mean mu=12.2

Let sigma be the standard deviation

Then P(X<12)=0.05 give P((X-mu)/sigma<(12-12.2)/sigma))=0.05

Hence,P(Z<z0)=0.05 where z0=(12-12.2)/sigma = -0.2/sigma

From excel function =normsinv(0.05) we get z0=-1.6448

So, -0.2/sigma=-1.6448

Hence sigma =0.2/1.6448=0.1216

Question-1

Part-a

Area under the curve Z<1.96=P(Z<1.96)=0.975 using excel function =normsdist(1.96)

Part-b

Area to the right of z=-0.79 is=P(Z>-0.79)

=1-P(Z<-0.79)

=1-0.2148 using excel function =normsdist(-0.79)

=0.7852

Part-c

Area between z=-2.45 and z=-1.32 is=P(-2.45<Z<-1.32)

=P(Z<-1.32)-P(Z<-2.45)

=0.0934-0.0071 using excel function =normsdist(-1.32) and =normsdist(-2.45)

=0.0863

Part-d

Area to the left of z=-1.39

=P(Z<-1.39)

=0.0823 using excel function =normsdist(-1.39)

Part-e

Area to the right of Z=1.96

=P(Z>1.96)

=1-P(Z<1.96)

=1-0.975 using excel function =normsdist(1.96)

=0.025

Part-f

Area between z=-2.3 and z=1.74

=P(-2.3<Z<1.74)

=P(Z<1.74)-P(Z<-2.3)

=0.9591-0.0107 using excel functions =NORMSDIST(1.74) and =NORMSDIST(-2.3)

=0.9483

Question-2

Part-a

We have to find z0 such that P(Z>z0)=0.7389

Or P(Z<z0)=1-0.7389= 0.2611     

We get z0=-0.6400 using excel function =NORMSINV(0.2611)

Part-b

We have to find z0 such that P(Z<z0)=0.2431

We get z0=-0.6964 using excel function =NORMSINV(0.2431)

Question-3

We have to find P(55<X<72)

=P(X<72)-P(X<55)

= 0.8849-0.3085 using excel functions =NORMDIST(72,60,10,TRUE) and =NORMDIST(55,60,10,TRUE)

=0.5764

Question-4

We have to find P(X>362)

=1-P(X<362)

=1-0.8925            using excel functions =NORMDIST(362,300,50,TRUE)

=0.1075

Question-5

We have to find P(X<2.3)= 0.0808 using excel function =NORMDIST(2.3,3,0.5,TRUE)

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.