Binomial probability distributions depend on the number of trials n of a binomial experiment and the probability of success p on
each trial. Under what conditions is it appropriate to use a normal approximation to the binomial? (Select all that apply.)nq > 10 np > 5 p > 0.5 np > 10 p < 0.5 nq > 5What is the probability of "12" or fewer successes? Use the normal approximation of the binomial distribution to answer this question. (Round your answer to four decimal places.)
1 answer:
The question is incomplete! Complete question along with answers and step by step explanation is provided below.
Question:
(a) Binomial probability distributions depend on the number of trials n of a binomial experiment and the probability of success p on each trial. Under what conditions is it appropriate to use a normal approximation to the binomial? (Select all that apply.)
nq > 10
np > 5
p > 0.5
np > 10
p < 0.5
nq > 5
(b) What is the probability of "12" or fewer successes for a binomial experiment with 20 trials. The probability of success on a single trial is 0.50. Use the normal approximation of the binomial distribution to answer this question. (Round your answer to four decimal places.)
Answer:
(a) The correct options are np > 5 and nq > 5
(b) P(x ≤ 12) = 0.8133
Step-by-step explanation:
Please refer to the attached images for explanation, I am unable to type in text editor due to some technical error!
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Answer:
1) The expected value of the "winnings" is $(-0.97)
2) The variance for the "winnings" is $0.57966
3) The standard deviation for the "winnings" is$0.761354
4) The game is not a fair game because one is expected to lose $0.97
Step-by-step explanation:
1) The probability of having a sum of 2 = 1/6×1/6 = 1/36
The probability of having a sum of 3 = 1/6×1/6 = 1/36
The probability of having a sum of 4 = 1/6×1/6 + 1/6×1/6 = 1/18
The probability of having a sum of 5 = 1/6×1/6 + 1/6×1/6 = 1/18
The probability of having a sum of 6 = 1/6×1/6 + 1/6×1/6 + 1/6×1/6 = 1/12
The probability of having a sum of 7 = 1/6×1/6 + 1/6×1/6 + 1/6×1/6 = 1/12
The probability of having a sum of 8 = 1/6×1/6 + 1/6×1/6 + 1/6×1/6 = 1/12
The probability of having a sum of 9 = 1/6×1/6 + 1/6×1/6 = 1/18
The probability of having a sum of 10 = 1/6×1/6 + 1/6×1/6 = 1/18
The probability of having a sum of 11 = 1/6×1/6 = 1/36
The probability of having a sum of 12 = 1/6×1/6 = 1/36
The values are;
For 2, we have 1/36 × (20 - 5) = 0.41
For 3, we have 1/36 × (20 - 5) = 0.41
For 4, we have 1/18 × (20 - 5) = 0.8
For 5, we have 1/18 × (10 - 5) = 0.2
For 6, we have 1/12 × (10 - 5) = 0.41
For 7, we have 1/12 × (10 - 5) = 0.41
For 8, we have 1/12 × (10 - 5) = 0.41
For 9, we have 1/18 × (-20 - 5) = -1.3
For 10, we have 1/18 × (-20 - 5) = -1.3
For 11, we have 1/36 × (-20 - 5) = -0.69
For 12, we have 1/36 × (-25 - 5) = -0.69
The expected value of the winnings is given as follows;
0.41 + 0.41
+ 0.8
+ 0.8
+ 0.8
+ 0.41
+ 0.41
+ -1.3
-1.3 - 0.69
- 0.69
= -0.97
Therefore, the expected value = $-0.97, which is one is expected to lose $0.97
2) Using Microsoft Excel, we have;
The variance for the "winnings", σ² = $0.57966
3) The standard deviation for the "winnings" = √σ² = √(0.57966) ≈ $0.761354
4) The game is not a fair game because one is expected to lose $0.97