The median of the following observations arranged in the ascending order is `42 `. Find x :
`22,24,33,37,x+1,x+3,44,47,51,58`.
Also find the upper quartile

Answer

To solve the problem, we need to find the value of x in the given data set and then determine the upper quartile. The data set is:

`22, 24, 33, 37, x + 1, x + 3, 44, 47, 51, 58`

### Step 1: Determine the number of observations
The number of observations (N) is 10.

### Step 2: Identify the median formula for an even number of observations
Since N is even (10), the median is calculated using the formula:

\[ \text{Median} = \frac{\text{(N/2)th term} + \text{(N/2 + 1)th term}}{2} \]

### Step 3: Substitute values into the median formula
Substituting N = 10 into the formula:

\[ \text{Median} = \frac{\text{(10/2)th term} + \text{(10/2 + 1)th term}}{2} = \frac{\text{5th term} + \text{6th term}}{2} \]

### Step 4: Identify the 5th and 6th terms in the data set
From the given data, the 5th term is \(x + 1\) and the 6th term is \(x + 3\).

### Step 5: Set up the equation for the median
We know the median is given as 42, so we can set up the equation:

\[ 42 = \frac{(x + 1) + (x + 3)}{2} \]

### Step 6: Simplify the equation
Multiply both sides by 2:

\[ 84 = (x + 1) + (x + 3) \]

Combine like terms:

\[ 84 = 2x + 4 \]

### Step 7: Solve for x
Subtract 4 from both sides:

\[ 80 = 2x \]

Divide by 2:

\[ x = 40 \]

### Step 8: Find the upper quartile
The upper quartile (Q3) is given by the formula:

\[ Q3 = \frac{3}{4}(N + 1) \text{th term} \]

### Step 9: Substitute N into the upper quartile formula
Substituting N = 10:

\[ Q3 = \frac{3}{4}(10 + 1) = \frac{3}{4} \times 11 = 8.25 \text{th term} \]

### Step 10: Determine the 8th and 9th terms
From the ordered data set, the 8th term is 47 and the 9th term is 51.

### Step 11: Calculate the upper quartile
To find Q3, we take the 8th term and add 0.25 of the difference between the 9th and 8th terms:

\[ Q3 = 47 + 0.25(51 - 47) = 47 + 0.25 \times 4 = 47 + 1 = 48 \]

### Final Answer
The value of \(x\) is 40, and the upper quartile \(Q3\) is 48.

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