1. a. On the basis of given data for the life in hours of 48 100-watt light bulbs, the frequency table can be constructed as shown in Table 1. Thebin values show the upper end of the frequency intervals or groups with group range as 50 hours. Table 1.1: Frequency table Bin Frequency 650 0 700 2 750 1 800 1 850 4 900 12 950 13 1000 6 1050 5 1100 4 More 0 b. On the basis of this frequency distribution, a histogram can be plotted using excel. The same is shown in Figure 1.1. Figure 1.1: Histogram c. It can be noticed that 25 out of 48 bulbs have a life in the middle range of 850-950 hours.
The data shows central tendency. The frequency polygon plot is very similar to that of a normal distribution. 2. a. The given data compares the sales of different models of Ford for the month of March over the years 2009 and 2010. Such a comparison can be best represented through a bar graph comparing the sales of each model across the two years. The visual representation for the same is shown in Figure 2.1. Figure 2.1: Visual representation of sales b. From the data and the graph above, it can be clearly seen that the sales in March 2010 have exceeded those in March 2009 for each and every model.
Also, there has been a big jump in the sales of Ranger over the year while for other models, the rise has been normal. 3. a. The arithmetic mean, median and the two quartiles for the given data are listed in the table 3.1. Table 3.1: Mean, Median & Quartiles Mean 567.24 Median 507 1st Quartile 407 3rd Quartile 761 b. The information on mean and median would help the manufacturer in estimating the probable life of his batteries. The same can be used as a customer specification.
The two quartiles would help to establish the most probable range of life untill failure. This would help in eliminating extreme values of range. c. The standard deviation, range and inter-quartile range for the given data are listed in Table 3.2. Table 3.2: Standard Deviation, Range & Inter-quartile range Std Deviation 239.8323 Range 843 Inter-quartile range 354 d. The manufacturer is quite reasonable in his claim that the batteries produced by the company last 400 hours. It can be seen that both the mean and median are quite large as compared to this value.
Also, the first quartile lies at 407 hours. This implies that only 25% of batteries have a life lower than 407 hours. Hence, the manufacturer is quite accurate in his advertising statement. In fact, he would do better if he increases his claim value. e. The decision on guarantee and monetary payments or gifts is entirely at the discretion of the company. If the company can afford to pay to 25% of its customers, the first quartile (25 percentile) value would suffice. However, the company would not ideally like to pay to more than 5-10% of the customers.
Hence, a 90th or 95th percentile of the data can be established accordingly and a guarantee can be given at that value. 4. a. The Mean and standard deviation of the given data can be calculated as follows: Table 4.1: Mean & Standard Deviation Waiting time in minutes Average Waiting Time for the range Frequency Total waiting time Second moment 10 but less than 15 12.5 7 87.5 1093.75 15 but less than 20 17.5 14 245 4287.5 20 but less than 25 22.5 23 517.5 11643.75 25 but less than 30 27.5 13 357.5 9831.25 30 but less than 35 32.5 15 487.5 15843.75 35 but less than 40 37.5 8 300 11250 Second moment 53950 Variance 53328.12 Mean 24.9375 Std Deviation 230.9288 b.
The median for the given distribution is calculated as: Median = L + (n/2 – CF)/F * i = 20 + (49 – 27)/26 * 5 = 24.231 The Ogive for the data is shown in Figure 4.1. Figure 4.1: Ogive c. The new mean and standard deviation with increased waiting times are calculated as: Table 4.2: New mean & standard deviation New Average Waiting Time Frequency Total waiting time Second moment 15 7 105 1575 21 14 294 6174 27 23 621 16767 33 13 429 14157 39 15 585 22815 45 8 360 16200 Second moment 77688 Variance 76792.49 Mean 29.925 Std Deviation 277.1146 d. The increase in waiting times of each customer by 20% increases both the mean and standard deviation considerably. e. In order to improve service, the managers can take orders over the phone or internet to book them in advance and prevent long queues.
New seating space may be acquired if capital investment is viable. 5. a. For the given survey, stratified sampling may work out to be the best because tourists may belong to various geographies and demographic profiles. It is important to consider all these subdivisions to ensure randomness and inclusiveness in the sample. b. The survey questionnaire can be developed as follows: i. What are your monthly savings? (Closed ended question) $2500 ii. ‘The current services and the fun activities for tourists in West Midlands are up to the mark’.
Do you Completely agree Partially agree Neutral Partially disagree Completely disagree with this statement? (multiple choice question with an attitude scale) iii. Please give three suggestions to improve the current services in West Midlands for tourists? (Open question) iv. How would you rate West Midlands as a tourist destination on a scale of 1 to 10? (Scaled response question) c. The personal information to be collected includes income level, level of savings, age, gender and address. d. The difficulties faced in the survey may be the lack of interest among the tourist to answer the survey questions.
Also the recency factor may come into play in their answers. The past experiences of tourists on similar trips may lead to several biases. e. For enhancing the findings of questionnaire survey, the responses to the questions can be analyzed using advanced statistical techniques such as regression analysis and anova. This would help in deriving appropriate conclusion on the basis of tourists’ responses.