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Calculate the 3rd quartile, 5th decile, and 90th percentile for the given grouped data.

Mathematics

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Answers

3 parts
a

69.5\text{69.5}

b

59.5\text{59.5}

c

79.5\text{79.5}

Step-by-step solution

9 steps

To calculate the 3rd quartile, 5th decile, and 90th percentile for grouped data, we first need to determine the true class boundaries and cumulative frequencies.

The given data is:

Score IntervalNumber of Students (ff)
30-395
40-498
50-597
60-6910
70-796
80-894

Step 1: Create a table with true class boundaries and cumulative frequencies.
The class width (ww) is 39.5−29.5=1039.5 - 29.5 = 10.
The total number of students (NN) is the sum of frequencies: 5+8+7+10+6+4=405+8+7+10+6+4 = 40.

Score IntervalTrue Class BoundariesFrequency (ff)Cumulative Frequency (cfcf)
30-3929.5 - 39.555
40-4939.5 - 49.5813
50-5949.5 - 59.5720
60-6959.5 - 69.51030
70-7969.5 - 79.5636
80-8979.5 - 89.5440

The formula for the kk-th percentile (PkP_k) is:

Pk=L+(kN100−cfbfp)×wP_k = L + \left(\frac{\frac{kN}{100} - cf_b}{f_p}\right) \times w

Where:
LL = lower class boundary of the percentile class
NN = total number of observations
cfbcf_b = cumulative frequency of the class before the percentile class
fpf_p = frequency of the percentile class
ww = class width

Part a· 3 steps

3rd quartile (Q3Q_3)

  1. Determine the position of Q3Q_3.

    The 3rd quartile is the 75th percentile (P75P_{75}).
    Position =3N4=3×404=30= \frac{3N}{4} = \frac{3 \times 40}{4} = 30.

  2. Identify the 3rd quartile class.

    The 30th observation falls in the class with a cumulative frequency of 30, which is 60-69 (true boundaries 59.5 - 69.5).
    L=59.5L = 59.5
    cfb=20cf_b = 20 (cumulative frequency of the class before 60-69)
    fp=10f_p = 10 (frequency of the class 60-69)
    w=10w = 10

  3. Calculate Q3Q_3.

    Q3=59.5+(30−2010)×10Q_3 = 59.5 + \left(\frac{30 - 20}{10}\right) \times 10 Q3=59.5+(1010)×10Q_3 = 59.5 + \left(\frac{10}{10}\right) \times 10 Q3=59.5+1×10Q_3 = 59.5 + 1 \times 10 Q3=59.5+10Q_3 = 59.5 + 10 Q3=69.5Q_3 = 69.5

    The 3rd quartile is 69.5\boxed{\text{69.5}}.

Part b

59.5\text{59.5}

5th decile (D5D_5)

  1. Determine the position of D5D_5.

    The 5th decile is the 50th percentile (P50P_{50}).
    Position =5N10=5×4010=20= \frac{5N}{10} = \frac{5 \times 40}{10} = 20.

  2. Identify the 5th decile class.

    The 20th observation falls in the class with a cumulative frequency of 20, which is 50-59 (true boundaries 49.5 - 59.5).
    L=49.5L = 49.5
    cfb=13cf_b = 13 (cumulative frequency of the class before 50-59)
    fp=7f_p = 7 (frequency of the class 50-59)
    w=10w = 10

  3. Calculate D5D_5.

    D5=49.5+(20−137)×10D_5 = 49.5 + \left(\frac{20 - 13}{7}\right) \times 10 D5=49.5+(77)×10D_5 = 49.5 + \left(\frac{7}{7}\right) \times 10 D5=49.5+1×10D_5 = 49.5 + 1 \times 10 D5=49.5+10D_5 = 49.5 + 10 D5=59.5D_5 = 59.5

    The 5th decile is 59.5\boxed{\text{59.5}}.

Part c

79.5\text{79.5}

90th percentile (P90P_{90})

  1. Determine the position of P90P_{90}.

    Position =90N100=90×40100=36= \frac{90N}{100} = \frac{90 \times 40}{100} = 36.

  2. Identify the 90th percentile class.

    The 36th observation falls in the class with a cumulative frequency of 36, which is 70-79 (true boundaries 69.5 - 79.5).
    L=69.5L = 69.5
    cfb=30cf_b = 30 (cumulative frequency of the class before 70-79)
    fp=6f_p = 6 (frequency of the class 70-79)
    w=10w = 10

  3. Calculate P90P_{90}.

    P90=69.5+(36−306)×10P_{90} = 69.5 + \left(\frac{36 - 30}{6}\right) \times 10 P90=69.5+(66)×10P_{90} = 69.5 + \left(\frac{6}{6}\right) \times 10 P90=69.5+1×10P_{90} = 69.5 + 1 \times 10 P90=69.5+10P_{90} = 69.5 + 10 P90=79.5P_{90} = 79.5

    The 90th percentile is 79.5\boxed{\text{79.5}}.

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