Question

The test scores of 40 students are summarized in the frequency table below. What is the standard deviation, mean, and median?

Score students

50-59 14

60-69 6

70-79 5

80-89 9

90-99 6

Answer 1

Answer: The test scores of 40 students are summarized in the frequency table below:

Solution:

Class	Midpoint (M)	f	M*f	M​​​​​​2*f
50-59	54.5	14	763	41583.50
60-69	64.5	6	387	24961.50
70-79	74.5	5	372.5	27751.25
80-89	84.5	9	760.5	64262.25
90-99	94.5	6	567	53581.50
Total		40	2850	212140

Mean:

Mean x̄ = ∑(M*f)/∑f

Mean x̄ = 2850/40

Therefore,

Mean x̄ = 71.25

Median:

= value of (n/2)th observation

= value of (40/2)th observation

= value of 20th observation

From the column of cumulative frequency cf, we find that the 20th observation lies in the class 70-79.

The median class is 69.5-79.5.

Now,

L=lower boundary point of median class =69.5

n=Total frequency =40

cf=Cumulative frequency of the class preceding the median class =20

f=Frequency of the median class =5

c=class length of median class =10

Therefore,

Median, M = L+ ((n/2-cf)/f)*c

M = 69.5 + ((20-20)/5) * 10

M = 69.5 * 0

Therefore,

Median,M = 69.5

Standard deviation, s:

s = √∑((M​​​​​​2*f) - (n*x̄​​​​​2)/n-1)

s = √(212140 - 40(71.25)2/40-1)

s = √((212140 - 203062)/39)

s = 15.2564

Therefore, Standard deviation, s = 15.2564

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