1) Find the following exact Binomial probabilities using the function Binomdist

Question

1) Find the following exact Binomial probabilities using the function Binomdist and the approx. Normal probabilities using the function Normdist.

P(x = 20), n = 120, p = .4

P(x > 8), n = 20, p = .15

P(x < 14), n = 40, p = .35

P(x 18), n = 50, p = .8

P(x 6), n = 25, p = .6

2.) Construct a binomial distribution (table) using Excel functions for the following: Eight percent of people in the US eligible to donate blood actually do. You randomly section 12 eligible blood donors and ask them if they donate blood. Create a histogram of this probability distribution. Describe the histogram.

3.) Use Excel functions to find the following probabilities for Z where Z is distributed N(0,1).

P(z < 1.74)

P(z > -.58)

P(-1.4 < z < 2.1)

P(z < -1.6 or z > 1.6)

4.) Use Excel functions to find the Z-score(s) corresponding to the following

            

a) P79

b) Separates top 10%         

5.) Use Excel functions to find the following Normal distribution probablilites

a) P (X > 78) when X ~ N (65, 10)

b) P(20 < X < 40) when X ~ N (42, 12)

6.) Use Excel functions to find the following given the probability when X ~ N(45, 5)

a.) P78                                                b.) Q1

P(x = 20), n = 120, p = .4

P(x > 8), n = 20, p = .15

P(x < 14), n = 40, p = .35

P(x 18), n = 50, p = .8

P(x 6), n = 25, p = .6

Explanation / Answer

Binomial probabilities:

P(x = 20), n = 120, p = .4

=BINOMDIST(20,120,0.4,FALSE) =2.11636E-08 = 0.0000

P(x > 8), n = 20, p = .15

=1-P(X<=8)=1-=BINOMDIST(8,20,0.15,TRUE)=1-0.99867=0.00133

P(x < 14), n = 40, p = .35

=P(x<=13)==BINOMDIST(13,40,0.35,TRUE)=0.044076

P(x 18), n = 50, p = .8

=BINOMDIST(18,50,0.8,TRUE)=1.61414E-11=0.0000

P(x 6), n = 25, p = .6)

=1-P(x<=5) = 1-=BINOMDIST(5,25,0.6,TRUE)=1- 5.35897E-05=0.99946

approx. Normal probabilities:

P(x = 20), n = 120, p = .4

Here mean = np=120*0.4= 48 > 5 and Variance= npq=120.0.4*0.6=28.8>0

So Standard deviation = 5.366

These are both over 5, so we can use the continuity correction factor.

Note:

Continuity Correction Factor Table

If   P(X=n) use   P(n – 0.5 < X < n + 0.5)
If   P(X>n) use   P(X > n + 0.5)
If   P(Xn) use P(X < n + 0.5)
If P (X<n) use   P(X < n – 0.5)
If P(X n) use   P(X > n – 0.5)

P(x = 20)=P(20-0.5<x<20+0.5)

=P(19.5<x<20.5)

=P(x<20.5) - P(x<19.5)

=P(Z<20.5-48/5.366)-P(Z<19.5-48/5.366)

=P(Z<-4.566)-P(Z<-4.752)

=NORMSDIST(-4.566)-=NORMSDIST(-4.752)

=2.48559E-06 - 1.00636E-06

=1.47923E-06

P(x > 8), n = 20, p = .15

Here mean = np=20*0.15= 3 < 5 and Variance= npq=2.55<5 . We can not use Continuty correction as bothe np and npq less than 5.

So Standard deviation = 1.597

P(x > 8),=1-P(x<=7) =1-P(z<7-3/1.597)

=1-NORMSDIST(2.505)

=1-0.99387

=0.00613

P(x < 14), n = 40, p = .35

Here mean = np=40*0.35= 14 > 5 and Variance= npq=9.1>5 .

These are both over 5, so we can use the continuity correction factor.

Standard deviation = sqrt(9.1)=3.017

P(X<14)=p(x<14-0.5)=P(x<13.5)=P(x<=13.4)=P(Z<13.4-14/3.017)=P(Z<-0.1988)=NORMSDIST(-0.1988)=0.4212

P(x 18), n = 50, p = .8

Here mean = np=50*0.8= 40 > 5 and Variance= npq=8>5 .

These are both over 5, so we can use the continuity correction factor.

Standadrd Deviation=SQRT(8)=2.828

P(x 18)=P(x<=18.5) =P(Z<=18.5-40/2.828)=P(Z<=4.355)=0.99999

Post remaining questions saperaelty.

Hope this will be helpful. Thanks and God Bless you:)

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