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ASSIGNMENT QUESTION INSTRUCTION: ANSWER ALL QUESTIONS QUESTION 1 Physicians use pulse rate to assess the health of patients. A pulse rate that is abnormally high or low suggests that there might be some medical issues. For example, a pulse rate that is too high might indicate that the patient has an infection or is dehydrated. Consider pulse rate measurements (in beats per minute) obtained from a simple random sample of 40 males and another simple random sample of 40 females as shown in Table 1. Table 1 : Pulse Rates (beats per minute) of Males and Females Males 68 64 88 72



64 72 60 88



76 60 96 72



56 64 60 64



84 76 84 88



72 56 68 60



60 68 60 60



56 84 72 84



88 56 64 56



56 60 64 72



72 68 80 64



68 68 80 76



68 72 96 72



68 72 64 80



64 80 76 96



76 80 104 88



60 76 72 72



88 80 60 72



95 88 108 64



Females 76 72 88 60 a)



Prepare a frequency distribution table for the pulse rates of males and females separately. Take 10 as a class width for both distributions and 50 as a lower limit of the first class for the males; 60 as a lower limit of the first class for the females.



b)



Prepare a histogram for each of the pulse rates distribution. Describe the shape of each distribution.



c)



Draw a cumulative frequency polygon for each of the pulse rates distribution. From your graph in (b), estimate the median. [Total: 15 marks]



QUESTION 2 a)



b)



From the data in Table 1, for each of the pulse rate distribution, calculate the following: i.



mean



ii.



mode



iii.



median



iv.



standard deviation



Determine the value of Pearson’s Coefficient of Skewness (PCS) for each distribution. Comment on the skewness of the data. [Total: 15 marks]



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MUKA SURAT TAMAT / END OF PAGE



Answer 1ai) MALES (BEAT PER MINUTES)



CLASS INTERVALS 50 – 59 60 – 69 70 – 79 80 – 89 90 – 99



FREQUENCY 6 17 8 8 1



CLASS INTERVALS 60 – 69 70 – 79 80 – 89 90 – 99 100 – 109



FREQUENCY 12 13 10 3 2



Answer 1aii) FEMALES



Answer b i) Histogram for males



Answer b ii) Histogram for females



As we can see from the distribution of histogram for males above, we can describe that the highest frequency of the graph value represented at 17 which have the pulse rate measurement at class interval 60 - 69. Then, the lowest frequency reading value represented at 1 which have the pulse rate measurement at class interval 90 – 99. At the frequency reading 8, they have 2 class interval which at 70 – 79 and 80 – 89. Next, for the females, we can see that the highest frequency is reading 13, at the class interval 70 – 79. Then, the lowest frequency value represented at 2, which have class interval at 100 – 109.



Answer c i) Cumulative Frequency Polygon MALES



For the median(Yellow color line) we chose ½ of (40+1) which is 20.5, for Q1(Red color line) we used ½ of 20.5 which is 10.25 and for Q3 (Green color line) we added these together to get the ¾ point, so 20.5+10.25 = 30.75 (For GCSE Maths you would get away with 20, 10 and 30) The median is 68.0, Q1 is 61.0 and Q3 is 78.0. Therefore the Interquartile range is 78.5 – 61.0 = 17.5



FEMALES



CLASS INTERVALS 60 – 69 70 –INTERVALS 79 CLASS 8050 – 89 – 59 9060 – 99 – 69 10070 – 109 – 79 80 – 89 90 – 99



FREQUENCY 12 13 FREQUENCY 10 6 3 17 2 8 8 1



CUMULATIVE FEQUENCY 12 25 FEQUENCY CUMULATIVE 35 6 3823 4031 39 40



For the median(Red color line) we chose ½ of (40+1) which is 20.5, for Q1(pink color line) we used ½ of 20.5 which is 10.25 and for Q3(Yellow color line) we added these together to get the ¾ point, so 20.5+10.25 = 30.75 (For GCSE Maths you would get away with 20, 10 and 30) The median is 76.0 , Q1 is 67.5 and Q3 is 85.0 . Therefore the Interquartile range is 85.0 – 67.5 = 17.5



Answer question no. 2 2a) i,ii,iii, and iv SEE THE CALCULATION ON THE NEXT PAGES.



2b) The coefficient is usually positive when the distribution is positively skewed, and negative when it is negatively skewed.



SEE THE SOLUTION ON THE NEXT PAGES A normal distribution is a bell-shaped distribution of data where the mean, median and mode all coincide. A frequency curve showing a normal distribution would look like this:



In a normal distribution, approximately 68% of the values lie within one standard deviation of the mean and approximately 95% of the data lies within two standard deviations of the mean. If there are extreme values towards the positive end of a distribution, the distribution is said to be positively skewed. In a positively skewed distribution, the mean is greater than the mode. For example:



A negatively skewed distribution, on the other hand, has a mean which is less than the mode because of the presence of extreme values at the negative end of the distribution.