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Angelina_Jolie [31]
3 years ago
14

The following table shows the number of hours some students in two universities spend reading each week: School A 7 2 3 10 17 14

10 22 2 School B 9 10 16 18 20 15 17 18 14 Part A: Create a five-number summary and calculate the interquartile range for the two sets of data. (6 points) Part B: Are the box plots symmetric? Justify your answer. (4 points)

Mathematics
1 answer:
pychu [463]3 years ago
8 0
The value of Q_{1}, Q_{2}, and Q_{3} for each schools is given in the first picture below. 

Q_{1} is the lower quartile
Q_{2} is the median
Q_{3} is the upper quartile

School A:
Minimum value is 2
Maximum value is 22
The lower quartile is 2.5
The median is 10
The upper quartile is 15.5

School B:
Minimum value is 9
Maximum value is 20
The lower quartile is 12
The median is 16
The upper quartile is 18

The box plot for each school is shown in the second picture

Box plot for school A isn't symmetrical. The data tails on the right
Box plot for school B isn't symmetrical. The data tails on the left


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Answer:

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3 years ago
A factory buys 10% of its components from suppliers B and the rest from supplier C. It is known that 6% of the components it buy
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Answer:

The percentage of the components bought from supplier C that are faulty = 88%

Complete Question:

A factory buys 10% of its components from supplier A, 30% from supplier B,and the rest from supplier C. it is known that 6% of the components it buys are faulty. of the components bought from supplier A, 9% are faulty and of the components bought from supplier B, 3% are faulty.

find the percentage of the components bought from supplier C that are faulty ?

Step-by-step explanation:

Let x be the total components bought by the factory

Components supplied by A = 10% of x

Components supplied by B = 30% of x

The rest supplied by C:

Percentage supplied by C = 100-(30+10) = 100-40 = 60%

Components supplied by C = 60% of x

% of all faulty components = 6%

Amount of faulty component = 6% of x

faulty components from A= 9%

Amount of faulty component from A= 9% × (6% of x) = 0.0054x

faulty components from B= 3%

Amount of faulty component from A= 3% × (6% of x) = 0.0018x

Amount of faulty component from C = (6% of x) - (0.0054x+0.0018x)

= 0.06x - 0.0072x

= 0.0528x

Let the percentage of the components bought from supplier C that are faulty = y%

y% × 6% of x = 0.0528x

y% × 0.06x = 0.0528x

y% = 0.0528x/0.06x

y% = 0.88 = 88%

The percentage of the components bought from supplier C that are faulty = 88%

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