- CFA Exams
- CFA Level I Exam
- Topic 1. Quantitative Methods
- Learning Module 7. Estimation and Inference
- Subject 2. The Central Limit Theorem and Inference
CFA Practice Question
Daily travel expenses for H&J Employees have a mean of $250 and a standard deviation of $50. If the accounting department is considered a random sample of size n = 49 then the probability that the average daily expense for the accounting department is greater than $260 is ______ (to the nearest 0.1%).
A. 8.1%
B. 42.1%
C. 57.9%
Explanation: The sampling distribution, the distribution of x-bars, for n = 49 (a large sample, n > 30) is approximately normal with mean = 250 and standard deviation = 50/7 = 7.14. The z-score for 260 is (260 - 250)/7.14 = 1.40. The table value for 1.4, P(x-bar < 260), is 0.9192. P(x-bar > 260) = 1 - 0.9192 = 0.0808. So, the probability the 49-member accounting department has an average expense of more than $260 is 8.1% (to the nearest 0.1%).
User Contributed Comments 9
User | Comment |
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capitalpirate | without doing working, once u have worked out sd=7.14, 260-250=10 = 1.5sd, you know the probability will be less than 10%. |
Bigworm | How did sd come out to be 7.14? |
naomixs | Why did the sd get divided by 7? I understand that there are 7 days in a week, but the question already gave the DAILY mean and sd, so I don't understand why sd was divided by 7?? Anyone? |
nike | 7 is the square root of 49. the way to calculate sample error. |
RexIdo | Y is it 7.14 and not 7 |
Shroud | It is supposed to be 50 divided by the square root of 49, so 50/7 = 7.14 |
Marinov | Again, how are we supposed to guess that without the tables. With this example, we can guess it but what if the answers were close? |
farrahkame | why is it $260-$250 and not $250-$260? usually we subtract the hypothesized mean from the sample mean and not the other way around no? |
Aoprita | If this is a normal distribution, variance is equal to the population variance divided by the sample size. Standard deviation 7.14 is the square root of variance 51.02. |