Given a standard Normal Distribution, find the area under the curve which lies?

Question

Given a standard Normal Distribution, find the area under the curve which lies? a. to the left of z=1.96 b. to the right of z=-0.79 c. between z= -2.45 and z=-1.32 d. to the left of z=-1.39 e. to the right of z=1.96 f. between z=-2.3 and z=1.74 Find the value of 2 if the area under a standard normal curve? a. to the right of z is .7389 b. to the left of z is .2431 Given a Normal Distribution with mu = 60 and o = 10, find the probability that X assumes a value between 55 and 72? Given a Normal Distribution with mu = 300 and alpha = 50, find the probability that X assumes a value greater than 362? A certain type of storage battery lasts an average of 3.0 years, with a standard deviation of 0.5 year" Assuming the battery lives are normally distributed, find the probability that a given battery will last less than 2.3 years

Explanation / Answer

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)

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