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1. the marks scored in a form three maths exam were recorded as follows…

Question

  1. the marks scored in a form three maths exam were recorded as follows 69 70 72 40 52 60 22 31 78 53 28 67 63 54 57 48 47 56 55 62 75 38 37 44 62 64 58 39 45 48 65 50 85 46 47 57 35 34 58 64 62 37 41 42 36 54 82 48 53 57 56 72 56 48 44 55 78 59 50 45 a) make a grouped frequency table with classes 20—29,30—39,40—49, etc b) what is the modal distribution of the test c) calculate the mean of the data d) calculate the median mark

Explanation:

Step1: Make grouped frequency table

  • For class \(20 - 29\): Count numbers like \(22,28\). Frequency \(f_1=2\)
  • For class \(30 - 39\): Numbers \(31,38,37,39,35,34,37,36\). Frequency \(f_2 = 8\)
  • For class \(40 - 49\): \(40,44,45,48,46,47,48,44,48\). Frequency \(f_3=9\)
  • For class \(50 - 59\): \(52,53,54,57,56,55,58,50,57,53,57,56,55,59,50\). Frequency \(f_4 = 15\)
  • For class \(60 - 69\): \(69,70,60,67,63,62,64,62,65,62,64\). Frequency \(f_5=11\)
  • For class \(70 - 79\): \(72,78,75,72,78\). Frequency \(f_6=5\)
  • For class \(80 - 89\): \(85,82\). Frequency \(f_7=2\)

Step2: Find modal distribution

The class with highest frequency is \(50 - 59\) (frequency \(15\))

Step3: Calculate mean

  • Let mid - points of classes be \(x_i\). For \(20 - 29\), \(x_1 = 24.5\); \(30 - 39\), \(x_2=34.5\); \(40 - 49\), \(x_3 = 44.5\); \(50 - 59\), \(x_4=54.5\); \(60 - 69\), \(x_5=64.5\); \(70 - 79\), \(x_6=74.5\); \(80 - 89\), \(x_7=84.5\)
  • Total frequency \(n=\sum f_i=2 + 8+9+15+11+5+2=52\)
  • \(\sum f_ix_i=(2\times24.5)+(8\times34.5)+(9\times44.5)+(15\times54.5)+(11\times64.5)+(5\times74.5)+(2\times84.5)\)

\(=49+276+400.5+817.5+709.5+372.5+169\)
\(=2794\)

  • Mean \(\bar{x}=\frac{\sum f_ix_i}{n}=\frac{2794}{52}\approx53.73\)

Step4: Calculate median

  • \(n = 52\), \(\frac{n}{2}=26\)
  • Cumulative frequencies: \(CF_1 = 2\), \(CF_2=2 + 8=10\), \(CF_3=10 + 9=19\), \(CF_4=19+15 = 34\)
  • Median class is \(50 - 59\)
  • \(L = 50\), \(CF = 19\), \(f = 15\), \(w=10\)
  • Median \(M = L+\frac{\frac{n}{2}-CF}{f}\times w\)

\(=50+\frac{26 - 19}{15}\times10\)
\(=50+\frac{70}{15}\approx50 + 4.67=54.67\)

Answer:

a)

ClassFrequency
30 - 398
40 - 499
50 - 5915
60 - 6911
70 - 795
80 - 892

b) \(50 - 59\)

c) \(\approx53.73\)

d) \(\approx54.67\)