In an examination the following scores were a chief by a group of students:
draw a cumulative frequency graph of the data and use it to find:
a) The median examination mark.
b) How many students score less than 65 marks
c) How many students scored between 50 and 80 marks
d) How many students failed, given that the pass mark is 50
e) The credit mark, mark given that the top 25% of students were awarded credits
f) The interquartile range.
Answers
hope this helps you out
The median of cumulative frequency is 67.86 and the interquartile range is 32.86.
What is median?
The mean is "the middle number is sorter, ascending or descending, list of numbers and can be more descriptive of that data set than the average".
What is interquartile range?
The interquartile range is "measure of where the 'middle fifty' is in data set".
According to the question,
Score Mid value Frequency Cumulative frequency
10 ≤ x < 20 15 1 1
20≤x < 30 25 3 4
30≤x < 40 35 6 10
40≤x < 50 45 15 25
50≤x < 60 55 14 39
60≤x < 70 65 28 67
70≤x < 80 75 18 85
80≤x < 90 85 11 96
90≤x < 100 95 4 100
Uniform interval is 10.
a) The median of examination mark
Frequency = 1+3+6+15+14+28+18+11+4 = 100
Median = [tex]l+\frac{(\frac{N}{2} - m )}{f}[/tex] × c where l = lower limit of the median class, N = total frequency, m = cumulative frequency of the pre median class,
f = frequency of the median class, c = uniform class interval.
= 50 + [tex]\frac{\frac{100}{2} - 25 }{14}[/tex] × 10
= 50 + [tex]\frac{50 - 25}{14}[/tex] × 10
= 50 +[tex]\frac{25}{14}[/tex] × 10
= 50 + 1.786 × 10
= 50 + 17.86
= 67.86.
Thus, the median of examination mark is 67.86.
b) The number of students score less than 65 marks.
In order to find the number of students score less than 65 marks sum of all frequency below 65.
= 1 + 3 + 6 + 15 + 14
= 39
Thus, the number of students score less than 65 marks is 39.
c) The number of students scored between 50 and 80 marks
In order to find the number of students scored between 50 and 80 marks only if add frequency between 50 and 80 marks.
= 14 + 28 + 18 + 11
= 71
Thus, the number of students scored between 50 and 80 marks is 71.
d) The number of students failed, given that the pass mark is 50.
In order to find the number of students failed, given that the pass mark is 50 only if add frequency below 50.
= 1 + 3 + 6 + 15
= 25.
Thus, the number of students failed, given that the pass mark is 50 is 25.
f) The interquartile range
Interquartile range = Q₃ - Q₁ [In words, difference between upper quartile range and lower quartile range]
Q₁ = l₁ + [tex]\frac{\frac{N}{4} - m }{f}[/tex] × c where l₁ is the lower limit of the Q₁ class, m₁ is the cumulative frequency of the pre median class, f₁ is the Frequency of the Q₁ class, c is the class interval, N is the total frequency.
[tex]\frac{N}{4}[/tex] = [tex]\frac{100}{4} = 25[/tex]; [tex]\frac{3N}{4}[/tex] = 75
= 40 + [tex]\frac{25 - 25}{6}[/tex] * 10 = 40
Q₁ = 40
Q₃ = l₃ + [tex]\frac{3\frac{N}{4} - m3 }{f3}[/tex] × c
= 60 + [tex]\frac{75 - 39}{28}[/tex] × 10
= 60 + [tex]\frac{36}{28}[/tex] × 10
= 60 + 1.2857 × 10
= 60 + 12.857
= 72.857
Interquartile range = upper quartile range - lower quartile range
= 72.857 - 40
= 32.8571 ≈ 32.86.
Thus, interquartile range is 32.86.
Hence, a) The median of examination mark is 67.86.
b) The number of students score less than 65 marks is 39.
c) The number of students scored between 50 and 80 marks is 71.
d) The number of students failed, given that the pass mark is 50 below 50 is 25.
f) The interquartile range is 32.86.
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Answers
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Step-by-step explanation:
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Answers
Answer:
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Answers
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Answers
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Answers
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Answers
Answer:
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Answers
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Step-by-step explanation:
Answer:
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Probability and Statistics
Acellus
The probability distribution for a
random variable x is given in the table.
Help Resources
-10
-5
0
5
10
15
20
.20
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.05
.1
Probability
.25
.1
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Answers
The probability distribution for a random variable x is given in the table x > 5 is 0.5.
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The probability of all the events occurring need to be 1.
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The table
Now,
Using these facts, we can find the probability that x > 5 from the given table:
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The probability of x = 10 is 0.25, the probability of x = 15 is 0.1, and the probability of x = 20 is 0.15. Adding them together, we get:
P(x > 5) = P(x = 10) + P(x = 15) + P(x = 20)
P(x > 5) = 0.25 + 0.1 + 0.15
P(x > 5) = 0.5
Therefore, the probability will be 0.5.
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Answers
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Since we are given the information that Ms. Wilson walked ¾miles each day. To calculate the total distance that Ms. Wilson will walk for 4 days, we have to multiply the distance walked each day by 4.
This will be:
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Find the value of c such that lim f(x) exists.
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√x+c. x > 2
Answers
Answer:
c=7
Step-by-step explanation:
because the limit exists and this is a piecewise function, we can assume that f(2) is a removable discontinuity and that both equations should be the same.
we can equal them to each other, plug in 2 for x and solve for c.
root(x + c) = 4 - .5x
root(2 + c) = 4 - .5(2)
root(2 + c) = 3
2 + c = 9
c = 7