# Probability Question - Pls verify my answer

1. Apr 10, 2007

### ORACLE

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?

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

2. Apr 10, 2007

### Mystic998

There's a definite problem with the last answer.

3. Apr 10, 2007

### ORACLE

hi mystic998

can you explain a bit about the problem.

thanks

4. Apr 10, 2007

### Dick

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.

5. Apr 10, 2007

### HallsofIvy

Staff Emeritus
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).

Yes, if you calculate P(18) as 1- P(22) then their sum is P(22)+ 1- P(22)= 1!:rofl:
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.

6. Apr 10, 2007

### ORACLE

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

7. Apr 10, 2007

### HallsofIvy

Staff Emeritus
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.

8. Apr 10, 2007

### ORACLE

thank you prof
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