5. A binomial distribution has 100 trials (n = 100) with a probability of succes

Question

5. A binomial distribution has 100 trials (n = 100) with a probability of success of 0.25 (p=0.25). We would like to find the probability of 34 or more successes using the normal distribution to approximate the binomial. Applying the continuity correction factor, what z-score should be used?
A. 2.079
B. 2.194
C. 1.963
D. 0.25

6. An accounting firm is planning for the next tax preparation season. From last year's returns, the firm collects a systematic random sample of 100 filings. The 100 filings showed an average preparation time of 90 minutes with a standard deviation of 140 minutes. What is the probability that the mean completion time is between 1 and 2 hours, i.e., 60 and 120 minutes?
A. Approximately 1
B. 0.1664
C. 0.8336
D. 0.9676

Explanation / Answer

Solution : 5

   C. 1.963

Description :

     Let X = number of successes out of 100 trials
X~binomial(n = 100, p = 0.25)

n = 100 > 30
np = (100)(0.25) = 25 > 5
n(1 - p) = (100)(0.75) = 75 > 5

Thus, we can use normal approximation.

X~normal(np = 25, np(1 - p) = 18.75) approximately

Z-score used
= (34 - 25 - 0.5) / sqrt(18.75)
= 1.962990915 ~ 1.963
Hence, Z-score should be used = 1.963

Solution : 6

   D. 0.9676

Description :

For distribution of ‘Means’, mean remains same but

Standard error of mean = 140/sqrt(100) = 14

Z for 60 = (60-90)/14 = -2.143

Z for 120 = (120-90)/14 = 2.143

P(-2.143<z<2.143) = 0.96

Hence,  the probability that the mean completion time is between 1 and 2 hours, i.e., 60 and 120 minutes is 0.96

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