Thread: Statistics: Approximating binomial distrubution

Help the problem is due in the morning.

I did post this in the urgent math help section, but I did not get an answer, so I am trying here.
I am going to try and delete my post in the Urgent section so I don't have duplicates.
Thanks all you math geniuses!


Use the binomial distriubution with n=20, p = 0.5 to compare accuracy of the normal approximation to the binomial.

a. Compute the exact probablities and corresponding normal approximations for y<2.

b. The normal approximation can be improved slightly by taking P(y<2.5).
Compare your results. Why should this help?

c. Compute the exact probabilities and corresponding normal approximations for 3<y<6 (note: 3.5<y<6.5 seems to work better for the normal approximation.)

Note: (0.5) to the 20th = 0.000001)

Originally Posted by Logicalmind

Help the problem is due in the morning.

I did post this in the urgent math help section, but I did not get an answer, so I am trying here.
I am going to try and delete my post in the Urgent section so I don't have duplicates.
Thanks all you math geniuses!


Use the binomial distriubution with n=20, p = 0.5 to compare accuracy of the normal approximation to the binomial.

a. Compute the exact probablities and corresponding normal approximations for y<2.

b. The normal approximation can be improved slightly by taking P(y<2.5).
Compare your results. Why should this help?

c. Compute the exact probabilities and corresponding normal approximations for 3<y<6 (note: 3.5<y<6.5 seems to work better for the normal approximation.)


Note: (0.5) to the 20th = 0.000001)

I would have thought that this is a straight forward application of the basic formulae. What have you done so far and where do you get stuck?

I did the first part calculating the exact propabilities using the binomial probability disribution.
P(y<2) = P(y=1) + P(y=0)
P(1) = [20!/1!(20-1)!] (0.5)^1 (0.5)^19 = 0.00002

P(0) = [20!/0!(20-0)!] (0.5)^0 (0.5)^20 = 0.0000005

0.00002+0.0000005 = 0.0000205

Then the normal approximation for P(y<2).
P(y<2) = P(y less than or equal to 1)
convert to the standard normal z value using the correction for continuity

z = [(a+0.5) - mu]/sigma
mu = np = 20(0.5) = 10
sigma = sqre[npq] = sqre[20(0.5)(0.5)] = 2.236

z = [(1+0.5) - 10]/2.236 = -3.8
using normal curve table, my table only goes up to 3.09 so I guessed at what 3.8 would be, about 0.4998
0.5 - 0.4998 = 0.0002
P(y<2) = P(y less than or equal to 1) = P(z less than or equal to -3.8) = 0.0002

Part (b) is where I get really stumped, my textbook says to "express the binomial probability to be approximated in the form P(x less than or equal to a)." I have no idea how to do this with a decimal number.
Does P(y<2.5) = P(y less than or equal to 1.5) ??? This doesn't seem to make any sense.

Also with part (c) I don't know how to express P(3.5<y<6.5) in the correct form.
Maybe P(y less than or equal to 5.5) - P(y less than or equal to 1.5)
I'm confused, let me know what I'm doing wrong.
Thanks

Originally Posted by Logicalmind

I did the first part calculating the exact propabilities using the binomial probability disribution.
P(y<2) = P(y=1) + P(y=0)
P(1) = [20!/1!(20-1)!] (0.5)^1 (0.5)^19 = 0.00002

P(0) = [20!/0!(20-0)!] (0.5)^0 (0.5)^20 = 0.0000005

0.00002+0.0000005 = 0.0000205 Mr F says: Using my TI-89 I get 0.0000200272.

Then the normal approximation for P(y<2).
P(y<2) = P(y less than or equal to 1)
convert to the standard normal z value using the correction for continuity

z = [(a+0.5) - mu]/sigma
mu = np = 20(0.5) = 10
sigma = sqre[npq] = sqre[20(0.5)(0.5)] = 2.236

z = [(1+0.5) - 10]/2.236 = -3.8
using normal curve table, my table only goes up to 3.09 so I guessed at what 3.8 would be, about 0.4998
0.5 - 0.4998 = 0.0002
P(y<2) = P(y less than or equal to 1) = P(z less than or equal to -3.8) = 0.0002 Mr F says: Your tables should go up to 3.8 .... But the values might not be too reliable. Using my TI-89 I get 0.00007237.

Part (b) is where I get really stumped, my textbook says to "express the binomial probability to be approximated in the form P(x less than or equal to a)." I have no idea how to do this with a decimal number.
Does P(y<2.5) = P(y less than or equal to 1.5) ??? This doesn't seem to make any sense. Mr F says: It's not immediately clear to me why the normal approximation to Pr(y < 2) is improved by using Pr(y<2.5) rather than Pr(y < 1.5) .... Perhaps if you draw the binomial histogram and overlay the normal distribution onto it .... Maybe you could play around with this http://www.ruf.rice.edu/~lane/stat_sim/binom_demo.html.

Also with part (c) I don't know how to express P(3.5<y<6.5) in the correct form.
Maybe P(y less than or equal to 5.5) - P(y less than or equal to 1.5)
I'm confused, let me know what I'm doing wrong.
Thanks

(c) P(3.5<y<6.5) = P(y < 6.5) - P(y < 3.5). (You're already using the normal approximation with these values so you just go ahead and calculate using the normal distribution).