Given that the median is 170, the observations are arranged as:
125, a, b, 170, 190, 210, 230
The mean deviation about the median is given by:
\[ \frac{0 + |45| + |60| + |20| + |40| + |170 - a| + |170 - b|}{7} = \frac{205}{7} \]
From this, we find:
\(|170 - a| + |170 - b| = 300 \implies a + b = 300\)
Now, the mean of the observations is:
\[ \text{Mean} = \frac{125 + a + b + 170 + 190 + 210 + 230}{7} = \frac{125 + 300 + 170 + 190 + 210 + 230}{7} = 175 \]
The mean deviation about the mean is:
\[ \frac{|125 - 175| + |a - 175| + |b - 175| + |170 - 175| + |190 - 175| + |210 - 175| + |230 - 175|}{7} \]
Simplifying:
\[ \frac{50 + |a - 175| + |b - 175| + 5 + 15 + 35 + 55}{7} = 30 \]
Class | 0 – 15 | 15 – 30 | 30 – 45 | 45 – 60 | 60 – 75 | 75 – 90 |
---|---|---|---|---|---|---|
Frequency | 11 | 8 | 15 | 7 | 10 | 9 |
Variance of the following discrete frequency distribution is:
\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Class Interval} & 0-2 & 2-4 & 4-6 & 6-8 & 8-10 \\ \hline \text{Frequency (}f_i\text{)} & 2 & 3 & 5 & 3 & 2 \\ \hline \end{array} \]
Let $ f(x) = \begin{cases} (1+ax)^{1/x} & , x<0 \\1+b & , x = 0 \\\frac{(x+4)^{1/2} - 2}{(x+c)^{1/3} - 2} & , x>0 \end{cases} $ be continuous at x = 0. Then $ e^a bc $ is equal to
Total number of nucleophiles from the following is: \(\text{NH}_3, PhSH, (H_3C_2S)_2, H_2C = CH_2, OH−, H_3O+, (CH_3)_2CO, NCH_3\)