2 a test administered to a statistic class of 50 students resulted in the ff scores\n45 65 33 72 50 36 43…

2 a test administered to a statistic class of 50 students resulted in the ff scores\n45 65 33 72 50 36 43 52\n60 52 32 63 52 82 46 78\n40 47 92 70 63 38 92 45\n41 36 86 46 77 36 34 46\na construct a frequency distribution using 10 class intervals with the lower limit of the class interval as the lowest score of the above data\ninclude\na. class boundaries\nb. class midpoint\nc. more than cumulative frequency less than cumulative frequency\nd. relative frequency\ne. percentage cumulative frequency

2 a test administered to a statistic class of 50 students resulted in the ff scores\n45 65 33 72 50 36 43 52\n60 52 32 63 52 82 46 78\n40 47 92 70 63 38 92 45\n41 36 86 46 77 36 34 46\na construct a frequency distribution using 10 class intervals with the lower limit of the class interval as the lowest score of the above data\ninclude\na. class boundaries\nb. class midpoint\nc. more than cumulative frequency less than cumulative frequency\nd. relative frequency\ne. percentage cumulative frequency

Answer

Explanation:

Step1: Find the range

The lowest score is 32 and the highest score is 92. Range = $92 - 32=60$.

Step2: Determine the class - width

Since we want 10 class intervals, class - width = $\frac{60}{10}=6$.

Step3: Define the class intervals

The first lower - limit is 32. The class intervals are: 32 - 37, 38 - 43, 44 - 49, 50 - 55, 56 - 61, 62 - 67, 68 - 73, 74 - 79, 80 - 85, 86 - 91, 92 - 97.

Step4: Calculate the frequencies

Count the number of scores in each class interval.

Step5: Calculate class boundaries

For the class interval 32 - 37, the lower class boundary is 31.5 and the upper class boundary is 37.5. Do this for all class intervals.

Step6: Calculate class mid - points

For the class interval 32 - 37, mid - point=$\frac{32 + 37}{2}=34.5$. Do this for all class intervals.

Step7: Calculate less than cumulative frequency

Start with the frequency of the first class and keep adding the frequencies of the subsequent classes.

Step8: Calculate more than cumulative frequency

Start with the total number of data points (50) and subtract the frequencies of the classes one by one.

Step9: Calculate relative frequency

Relative frequency of a class = $\frac{\text{Frequency of the class}}{\text{Total number of data points}}$.

Step10: Calculate percentage cumulative frequency

Percentage cumulative frequency of a class = $\frac{\text{Cumulative frequency of the class}}{\text{Total number of data points}}\times100$.

Class Interval Class Boundaries Class Mid - point Frequency Less than Cumulative Frequency More than Cumulative Frequency Relative Frequency Percentage Cumulative Frequency
32 - 37 31.5 - 37.5 34.5 7 7 50 $\frac{7}{50}=0.14$ $\frac{7}{50}\times100 = 14%$
38 - 43 37.5 - 43.5 40.5 6 13 43 $\frac{6}{50}=0.12$ $\frac{13}{50}\times100 = 26%$
44 - 49 43.5 - 49.5 46.5 10 23 37 $\frac{10}{50}=0.2$ $\frac{23}{50}\times100 = 46%$
50 - 55 49.5 - 55.5 52.5 5 28 27 $\frac{5}{50}=0.1$ $\frac{28}{50}\times100 = 56%$
56 - 61 55.5 - 61.5 58.5 2 30 22 $\frac{2}{50}=0.04$ $\frac{30}{50}\times100 = 60%$
62 - 67 61.5 - 67.5 64.5 4 34 20 $\frac{4}{50}=0.08$ $\frac{34}{50}\times100 = 68%$
68 - 73 67.5 - 73.5 70.5 3 37 16 $\frac{3}{50}=0.06$ $\frac{37}{50}\times100 = 74%$
74 - 79 73.5 - 79.5 76.5 3 40 13 $\frac{3}{50}=0.06$ $\frac{40}{50}\times100 = 80%$
80 - 85 80.5 - 85.5 82.5 2 42 10 $\frac{2}{50}=0.04$ $\frac{42}{50}\times100 = 84%$
86 - 91 85.5 - 91.5 88.5 3 45 8 $\frac{3}{50}=0.06$ $\frac{45}{50}\times100 = 90%$
92 - 97 91.5 - 97.5 94.5 5 50 5 $\frac{5}{50}=0.1$ $\frac{50}{50}\times100 = 100%$

Answer:

The frequency distribution table is shown above.