Page 260 - BA2 Integrated Workbook - Student 2017
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Fundamentals of Management Accounting




               11.3 Band D

                     The mode is the value which appears the most frequently, so 43 appears 3
                     times.


                     The median is the value of the middle item in a distribution. There are 3 marks
                     above 50 and 3 marks below, so the mode is 50.

                     The mean is the average = (50, 57, 43, 52, 43, 62, 43) ÷ 7 = 50.


               11.4 B

                                                                                    ଵ
                     There are 360° in a circle, so if wages represent 60° that is   of total cost:
                                                                                    ଺
                     ଵ  × $720,000 = $120,000
                     ଺



               CHAPTER 12 – RISK 2: PROBABILITY


               12.1 The complete sentences are:

                     Empirical probabilities can be calculated from samples of past observations.

                     Subjective probabilities are based on judgement.

                     Exact probabilities can be applied to the population of outcomes for a certain
                     event.


               12.2 The percentage of containers with contents over 500 ml is 26.43%.

                     We have a normal distribution with μ = 499.5 and σ = 0.8.


                     Calculating the z score, we get:
                                      500 – 499.5
                     z =             ——————             = 0.625
                                          0.8
                     Once we have calculated our ‘z score’ we can look this up on the normal
                     distribution table to find the area under the curve, which equates to the
                     percentage chance (probability) of that value occurring.


                     So if we have calculated a z score of 0.625. From the table the value is 0.2357

                     We want P(x > 500) = P(z > 0.625) = (0.5 – 0.2357) = 0.2643

                     Therefore, 26.43% of containers have contents over 500 ml.




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