Q.1 – Q.2 , Q.3 – Q.4 Q.5 , Q.6 , Q.7 , Q.9 , Q. 10 , Q. 11 Q.12 , Q.13 , Q. 14 , Q. 15 Q.16 to Q. 19
Answer 3
Group 1 = 20 values
Mean = 85 , Median = 80
Total of 20 values = 20 x 85 = 1700
Group 2 = 30 values
Mean = 75, Median = 72
Total of 30 Values = 30 x 75 = 2250
(a). Mean of 50 values = (1700 +2250)/50
Mean of 50 values = 79
(b). As there is no list of numbers, we cannot assume any number as a median for 50 values.
Answer 4
| Refer the Book for
Random Variable ‘X’ |
Refer the Book for Relative Frequency | Cumulative Frequency
(Add values of column ‘B’) |
||
| A | B | A x B | C = A X B | D |
| 0 x 0.18 | 0 | 0.18 | ||
| 1 x 0.33 | 0.33 | 0.33+0.18=0.51 | ||
| 2 x 0.10 | 0.2 | 0.51+0.1 =0.61 | ||
| 3 x 0.06 | 0.18 | 0.61+0.06 = 0.67 | ||
| 4 x 0.33 | 1.32 | 0.67+0.33 =1.0 | ||
Mean of Random Variable ‘X’ = Expected Value = Sum of Column ‘C’
=0 + 0.33 + 0.2 + 0.18 + 1.32
Mean = 2.03
The median is the central tendency which is unaffected by high or low values related to the rest of the data. Here median of a random variable ‘X’ is any number ‘m’ from the column ‘A’ such that
P ( X ≤ m ) ≥ 0.5 and P ( X ≥ m ) ≥ 0.5. This means that at least half the values of X are greater than or equal to m and at least half the values of X are less than or equal to m.
To find the median of random variable ‘X’ just check for cumulative values as per column ‘D’.
P(X < 1 ) = P(x=0) + P(x=1) = 0.18 + 0.33 = 0.51
P(X > 1 ) = P(x=1) + P(x=2) + P(x=3) + P(x=3) = 0.33 + 0.10+ 0.06 + 0.33 = 0.82
So median of random variable ‘X’ is 1.
Q.1 – Q.2 , Q.3 – Q.4 Q.5 , Q.6 , Q.7 , Q.9 , Q. 10 , Q. 11 Q.12 , Q.13 , Q. 14 , Q. 15 Q.16 to Q. 19
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