Question

#31 what is the first (i.e. lower) quartile: q1 = ?
(a) 6 (b) 7 (c) 8 (d) 9 (e) none of these
#32 what is the second (i.e. middle) quartile: q2 = ?
(a) 6 (b) 7 (c) 8 (d) 9 (e) none of these
#33 what is the third (i.e. upper) quartile: q3 = ?
(a) 6 (b) 7 (c) 8 (d) 9 (e) none of these
#34 what is the inter - quartile range: iqr = ?
(a) 1 (b) 2 (c) 3 (d) 4 (e) none of these

Explanation:

Since the data - set is not provided, we cannot calculate the quartiles and the inter - quartile range. But assuming we have a data - set \(x_1,x_2,\cdots,x_n\) arranged in ascending order:

Step1: Arrange data in ascending order

Let the data - set be \(x_1\leq x_2\leq\cdots\leq x_n\).

Step2: Calculate position of \(Q_1\)

If \(n\) is the number of data points, the position of \(Q_1\) is \(i_1=\frac{n + 1}{4}\) (for \(n\) odd) or \(i_1=\frac{1}{2}(\frac{n}{4}+\frac{n}{4}+ 1)\) (for \(n\) even). If \(i_1\) is an integer, \(Q_1=x_{i_1}\), if \(i_1\) is not an integer, say \(i_1 = k + r\) where \(k\) is the integer part and \(r\) is the fractional part, \(Q_1=(1 - r)x_k+rx_{k + 1}\).

Step3: Calculate position of \(Q_2\)

The position of \(Q_2\) (the median) is \(i_2=\frac{n + 1}{2}\) (for \(n\) odd) or \(i_2=\frac{1}{2}(\frac{n}{2}+\frac{n}{2}+1)\) (for \(n\) even). If \(i_2\) is an integer, \(Q_2=x_{i_2}\), if \(i_2\) is not an integer, \(Q_2=(1 - r)x_k+rx_{k + 1}\) where \(i_2=k + r\).

Step4: Calculate position of \(Q_3\)

The position of \(Q_3\) is \(i_3=\frac{3(n + 1)}{4}\) (for \(n\) odd) or \(i_3=\frac{1}{2}(\frac{3n}{4}+\frac{3n}{4}+1)\) (for \(n\) even). If \(i_3\) is an integer, \(Q_3=x_{i_3}\), if \(i_3\) is not an integer, \(Q_3=(1 - r)x_k+rx_{k + 1}\) where \(i_3=k + r\).

Step5: Calculate \(IQR\)

\(IQR = Q_3-Q_1\)

Since we have no data, we cannot give a definite answer for each question. But the general formulas for calculation are as above.

Answer:

[#31] E. None of these
[#32] E. None of these
[#33] E. None of these
[#34] E. None of these