If `overline (x_(2)) " and" overline (x_(2))` are the means of two distributions such that `overline (x_(1)) lt overline (x_(2)) "and" overline (x)` i

Question

If `overline (x_(2)) " and" overline (x_(2))` are the means of two distributions such that `overline (x_(1)) lt overline (x_(2)) "and" overline (x)` i

If `overline (x_(2)) " and" overline (x_(2))` are the means of two distributions such that `overline (x_(1)) lt overline (x_(2)) "and" overline (x)` is the mean of the combined distriubtion, then
A. `overline (x) lt overline (x_(1))`
B. `overline (x) gt overline (x_(2))`
C. `overline (x)=(overline (x_(1))+overline (x_(2)))/(2)`
D. `overline (x_(1)) lt overline (x) lt overline (x_(2))`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - D
Let `n_(1) " and" n_(2)` be the number of observations in two groups having means `overline(x_(1)) " and" overline (x_(2))`, respectively. Then,
`overline (x)=(n_(1)overline(x_(1))+n_(2)overline(x_(2)))/(n_(1)+n_(2))`
Now, `overline (x)-overline(x_(1))=(n_(1)overline(x_(1))+n_(2)overline(x_(2)))/(n_(1)-n_(2))-overline(x_(1))`
` =(n_(2)(overline(x_(2))-overline(x_(1))))/(n_(1)+n_(2)) gt 0 " " [because overline(x_(2))gt overline(x_(1))]`
`implies overline(x) gt overline(x_(1))`,
and `overline(x)-overline(x_(2))=(n(overline(x_(1))-overline(x_(2))))/(n_(1)+n_(2)) lt 0 " " [ because overline(x_(2)) gt overline(x_(1))]`
`implies overline (x) lt overline (x_(2))`
From (1) and (2), `overline(x_(1)) lt overline(x) lt overline(x_(2))`.