
#1
Apr1007, 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.(1p) = √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
Apr1007, 08:59 AM

P: 206

There's a definite problem with the last answer.




#3
Apr1007, 09:20 AM

P: 12

hi mystic998
can you explain a bit about the problem. thanks 



#4
Apr1007, 09:46 AM

Sci Advisor
HW Helper
Thanks
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 10.1915). You just need to practice using your table.




#5
Apr1007, 10:46 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

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
Apr1007, 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 



#7
Apr1007, 12:26 PM

Math
Emeritus
Sci Advisor
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.




#8
Apr1007, 12:30 PM

P: 12

thank you prof



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