Median is the middle value in a set of data when it is arranged in ascending or descending order. To find the median, we need to arrange the given data in ascending order.

Given data:
n = 230
l = 40
cf = 76
h = 10
f = 65

To find the exact values, we can use the cumulative frequency (cf) and the frequency (f) of each class interval.

First, let's calculate the lower limits (LL) and upper limits (UL) of each class interval using the formula:
LL = n + (l - cf) × h
UL = LL + h

| Class Interval | LL | UL | f |
|----------------|----|----|---|
| 30-40 | 230 + (40 - 76) × 10 = 230 + (-360) = -130 | -130 + 10 = -120 | 65 |
| 40-50 | -120 + 10 = -110 | -110 + 10 = -100 | 40 |
| 50-60 | -100 + 10 = -90 | -90 + 10 = -80 | 25 |
| 60-70 | -80 + 10 = -70 | -70 + 10 = -60 | 20 |
| 70-80 | -60 + 10 = -50 | -50 + 10 = -40 | 15 |
| 80-90 | -40 + 10 = -30 | -30 + 10 = -20 | 10 |
| 90-100 | -20 + 10 = -10 | -10 + 10 = 0 | 5 |

Now, let's calculate the cumulative frequencies (cf) for each class interval by adding up the frequencies:
| Class Interval | LL | UL | f | cf |
|----------------|----|----|----|----|
| 30-40 | -130 | -120 | 65 | 65 |
| 40-50 | -110 | -100 | 40 | 65 + 40 = 105 |
| 50-60 | -90 | -80 | 25 | 105 + 25 = 130 |
| 60-70 | -70 | -60 | 20 | 130 + 20 = 150 |
| 70-80 | -50 | -40 | 15 | 150 + 15 = 165 |
| 80-90 | -30 | -20 | 10 | 165 + 10 = 175 |
| 90-100 | -10 | 0 | 5 | 175 + 5 = 180 |

Now, let's find the median. Since the total frequency is 230, the median will be located in the class interval where the cumulative frequency (cf) is greater than or equal to 115 (half of the total frequency).

The cumulative frequency of the class interval 40-50 is 105, and the cumulative frequency of the class interval 50-60 is 130. Therefore, the median lies in the class interval 50-60.

To find the exact value of the median,

L+(n/2-cf)*h