TI-83 or TI-84: binompdf(n, p, r) is the same as BINOMDIST(r, n, p, 0) in Excel.
TI-83 or TI-84: binomcdf(n, p, r) is the same as BINOMDIST(r, n, p, 1) in Excel.
Problems:
a) What is the probability of 20 or less successes in 30 independent trials when the probability of success on any one trial is .6?
b) What is the probability of 20 or less successes in 30 independent trials when the probability of success on any one trial is .65?
c) What is the probability of 20 or less successes in 30 independent trials when the probability of success on any one trial is .7?
d) What is the probability of 30 or more successes in 40 independent trials when the probability of success on any one trial is .8?
e) What is the probability of 30 or more successes in 40 independent trials when the probability of success on any one trial is .75?
f) What is the probability of 30 or more successes in 40 independent trials when the probability of success on any one trial is .7?
g) Optional for those with TI-83 calculators or Excel. Find np and nq for each problem and how close the approximations are.
Answers in the comments.
1 comment:
a) What is the probability of 20 or less successes in 30 independent trials when the probability of success on any one trial is .6?
Answer: binomcdf(30, .6, 20) = .823714... rounds to .8237
z-score: (20 + .5 - 30*.6)/sqrt(.6*.4*30) =
.9316... ~= .93
.93 on the lookup table goes to .8238
b) What is the probability of 20 or less successes in 30 independent trials when the probability of success on any one trial is .65?
Answer: binomcdf(30, .65, 20) = .642459... rounds to .6425
z-score: (20 + .5 - 30*.65)/sqrt(.65*.35*30) =
.3827... ~= .38
.38 on the lookup table goes to .6480
c) What is the probability of 20 or less successes in 30 independent trials when the probability of success on any one trial is .7?
Answer: binomcdf(30, .7, 20) = .41119... rounds to .4112
z-score: (20 + .5 - 30*.7)/sqrt(.7*.3*30) = -0.199...
~= -.20
-.20 on the lookup table goes to .4207
d) What is the probability of 30 or more successes in 40 independent trials when the probability of success on any one trial is .8?
Answer: 1 - binomcdf(40, .8, 29) = .83923... rounds to .8392
z-score: (30 - .5 - 40*.8)/sqrt(.8*.2*40) = -0.988...
~= -.99
-.99 on the lookup table goes to .1611, 1-.1611 = .8389
e) What is the probability of 30 or more successes in 40 independent trials when the probability of success on any one trial is .75?
Answer: 1 - binomcdf(40, .75, 29) = .58390... rounds to .5839
z-score: (30 - .5 - 40*.75)/sqrt(.75*.25*40) = -0.18...
~= -.18
-.18 on the lookup table goes to .4286, 1-.4286 = .5714
f) What is the probability of 30 or more successes in 40 independent trials when the probability of success on any one trial is .7?
Answer: 1 - binomcdf(40, .7, 29) = .30874... rounds to .3087
z-score: (30 - .5 - 40*.7)/sqrt(.7*.3*40) = 0.517...
~= .52
.52 on the lookup table goes to .6985, 1-.6985 = .3015
g) The approximations get should better when both np and nq are greater than 10. Here are those values for all problems.
a) np = 18, nq = 12, approximation is .0001 away
b) np = 19.5, nq = 10.5, approximation is .0055 away
c) np = 21, nq = 9, approximation is .0095 away
d) np = 32, nq = 8, approximation is .0003 away
e) np = 30, nq = 10, approximation is .0125 away
f) np = 28, nq = 12, approximation is .0072 away
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