Solve these problems on charges by finding out Q1 and Q2. Q1Q2= 4x10^6, Q1 + Q2 =8.00×10^-6

Q_1 = 7Q

1

=7

Q_2 = 10Q

2

=10

Q_3 = 13.5Q

3

=13.5

Step-by-step explanation:

Given

5, 7, 7, 8, 10, 11, 12, 15, 17.

Required

Determine Q1, Q2 and Q3

The number of data is 9

Calculating Q1:

Q1 is calculated as:

Q_1 = \frac{1}{4}(N + 1)Q

1

=

4

1

(N+1)

Substitute 9 for N

Q_1 = \frac{1}{4}(9 + 1)Q

1

=

4

1

(9+1)

Q_1 = \frac{1}{4}*10Q

1

=

4

1

∗10

Q_1 = 2.5th\ itemQ

1

=2.5th item

This means that the Q1 is the mean of the 2nd and 3rd data.

So:

Q_1 = \frac{1}{2}(7+7)Q

1

=

2

1

(7+7)

Q_1 = \frac{1}{2}*14Q

1

=

2

1

∗14

Q_1 = 7Q

1

=7

Calculating Q2:

Q2 is calculated as:

Q_2 = \frac{1}{2}(N + 1)Q

2

=

2

1

(N+1)

Substitute 9 for N

Q_2 = \frac{1}{2}(9 + 1)Q

2

=

2

1

(9+1)

Q_2 = \frac{1}{2}*10Q

2

=

2

1

∗10

Q_2 = 5th\ itemQ

2

=5th item

Q_2 = 10Q

2

=10

Calculating Q3:

Q3 is calculated as:

Q_3 = \frac{3}{4}(N + 1)Q

3

=

4

3

(N+1)

Substitute 9 for N

Q_3 = \frac{3}{4}(9 + 1)Q

3

=

4

3

(9+1)

Q_3 = \frac{3}{4}*10Q

3

=

4

3

∗10

Q_3 = 7.5th\ itemQ

3

=7.5th item

This means that the Q3 is the mean of the 7th and 8th data.

So:

Q_3 = \frac{1}{2}(12+15)Q

3

=

2

1

(12+15)

Q_3 = \frac{1}{2}*27Q

3

=

2

1

∗27

Q_3 = 13.5Q

3

=13.5

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