Probability Question - Pls verify my answer


by ORACLE
Tags: probability, verify
ORACLE
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#1
Apr10-07, 08:57 AM
P: 12
hi,
pls comment on my solution.
thanks.


Question

A binomial probability distribution has p= 0.20 and n = 100.
a)What is the mean and standard deviation?
b)Can you approximate these Binomial probabilities by the normal probabilities? Explain.
c)What is the probability of more than 24 successes?
d)What is the probability of 18 to 22 successes?

Answers

a.) The formula to calculate the mean of Binomial Probability Distribution is
μ = n.p
= 100 X .2
= 20

The formula to calculate the Standard Deviation of a Binomial Probability Distribution is
σ = √np.(1-p)
= √100 X .2 X .8
= 4

b.) Since the binomial distribution approaches the normal distribution as the number of trials increases, we can use the normal distribution to approximate binomial probabilities. These values will all be approximations; we could use the binomial probability to obtain the actual answers, but in some cases these probabilities are difficult to calculate by hand.
We can use the standard normal distribution with the following conversion formula: Z = (X - np)/ sqrt(npq)

c.) Probability of more than 24 successes can be phrased as
P(X>24)
Z value of X = (X - μ)/ σ
= (24.5 20)/4
= 1.125 ~ 1.13
= 0.3708
.3708 is the probability between the mean (20) and 24.
P(X>24) = 0.5 - .3708 = 0.1292

d.) Probability of 18 to 22 successes can be formulated as
P(18<=X<=22)
Z value= (X - μ)/ σ
= (22 20)/4
= .5
P(22) = 0.1915
P(18)= (18 20)/4
= -.5
Z Value= 1 0.1915
= 0.8085
Therefore P(18<=X<=22) = P(22)+ P(18)
= .1915 + 8085
= 1
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Mystic998
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#2
Apr10-07, 08:59 AM
P: 206
There's a definite problem with the last answer.
ORACLE
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#3
Apr10-07, 09:20 AM
P: 12
hi mystic998

can you explain a bit about the problem.

thanks

Dick
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#4
Apr10-07, 09:46 AM
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P: 25,167

Probability Question - Pls verify my answer


The number you have pulled out of a table 0.1915 represents the probability that the variable falls between Z=0 and Z=0.5. So the probability that it falls between Z=-0.5 and Z=0 is also 0.1915, since the normal distribution is symmetric (NOT 1-0.1915). You just need to practice using your table.
HallsofIvy
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#5
Apr10-07, 10:46 AM
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PF Gold
P: 38,879
Quote Quote by ORACLE View Post
hi,
pls comment on my solution.
thanks.


Question

A binomial probability distribution has p= 0.20 and n = 100.
a)What is the mean and standard deviation?
b)Can you approximate these Binomial probabilities by the normal probabilities? Explain.
c)What is the probability of more than 24 successes?
d)What is the probability of 18 to 22 successes?

Answers

a.) The formula to calculate the mean of Binomial Probability Distribution is
μ = n.p
= 100 X .2
= 20

The formula to calculate the Standard Deviation of a Binomial Probability Distribution is
σ = √np.(1-p)
= √100 X .2 X .8
= 4

b.) Since the binomial distribution approaches the normal distribution as the number of trials increases, we can use the normal distribution to approximate binomial probabilities. These values will all be approximations; we could use the binomial probability to obtain the actual answers, but in some cases these probabilities are difficult to calculate by hand.
We can use the standard normal distribution with the following conversion formula: Z = (X - np)/ sqrt(npq)

c.) Probability of more than 24 successes can be phrased as
P(X>24)
Z value of X = (X - μ)/ σ
= (24.5 – 20)/4
= 1.125 ~ 1.13
= 0.3708
.3708 is the probability between the mean (20) and 24.
P(X>24) = 0.5 - .3708 = 0.1292

d.) Probability of 18 to 22 successes can be formulated as
P(18<=X<=22)
Z value= (X - μ)/ σ
= (22 – 20)/4
= .5
P(22) = 0.1915
P(18)= (18 – 20)/4
= -.5
Z Value= 1 – 0.1915
First, it is not P(18) that is (18- 20)/4, it is the z-value. You are looking for P(-.5). Most importantly, that is not "1- P(5)". I assume your table has only Z> 0 so you are using the symmetry of the normal distribution to get P for negative values. Your mistake is that P(0)= 1/2, not 1. P(-.5)= 1/2- P(.5).

= 0.8085
Therefore P(18<=X<=22) = P(22)+ P(18)
= .1915 + 8085
= 1
Yes, if you calculate P(18) as 1- P(22) then their sum is P(22)+ 1- P(22)= 1!
Even with the correct calculation, that P(18)= .5- P(22), their sum would be P(22)+ .5- P(22)= .5. P(18<=X<= 22)= P(22)- P(18), not the sum.
ORACLE
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#6
Apr10-07, 11:48 AM
P: 12
hi prof,

thanks. but now i get a -ve probability. what does it mean?
pls help.

P(18<=X<=22) = P(22) – P(18)
= .1915 - .3085
= -0.117
HallsofIvy
HallsofIvy is offline
#7
Apr10-07, 12:26 PM
Math
Emeritus
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Thanks
PF Gold
P: 38,879
Sorry, I should have recognized that if your table is giving you P(.5)= .1915, then it is giving you P(0<= Z<= .5). The probability of P(-infinity<= Z<= 0)= 0.5 itself so P(-infinity<= Z<= .5) is .5+ .1915= .6915. Then P(-infinity<= Z<= -.5) is, as I said, 1/2- .1915= .3085. P(-.5<= Z<= .5)= .6915- .3085= .383. If you look at that closely, you will see that .383= 2(.1915). Remember that the probability is the "area under the normal curve" and the normal curve is symmetric! The area from -.5 to .5 is just twice the area from 0 to .5.
ORACLE
ORACLE is offline
#8
Apr10-07, 12:30 PM
P: 12
thank you prof


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