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Hi Mr. Maffot,
Can you please explain how to answer these two questions…I have in fact watched your lectures. But I find it difficult to answer these two. Please help me answer these two questions.
Thanks.
Q1.) The percentage probability that a student will score more than 90 marks in an accounting exam is 2.28%. The marks scored from the accounting exam follow a normal distribution with a mean score of 70.
What is the standard deviation for the distribution?
Q2.) The number of daily complaints to a local government office has a deviation of 7 and the coefficient of variation is 20%.
What is the mean number of daily complaints?.
Why are you attempting questions for which you do not have answers? You should be using a Revision Kit from one of the ACCA Approved Publishers. They have answers and explanations.
1. If the probability of being more than 90 is 2.28%, then the probability of being between the mean of 70 and 90 is 50% – 2.28% = 47.72% (or 0.4772).
You can then look ‘backwards’ from the table to see how many standard deviations give a probability of 0.4772. If you divide 20 (90 – 70) by this number of standard deviations you will have the value of a standard deviation.2. The coefficient of variation is the standard deviation divided by the mean.
Therefore, rearranging, the mean is equal to the standard deviation (7) divided by the coefficient of variation (0.2).Thank you so much Mr. Maffot. I want to mention that this question is from a mock exam. Thanks again!!!!
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