- A34
- B66
- C21
- D27
Step-by-Step Solution
Add the frequencies of the classes lying below 30: 7 + 14 + 13 = 34 intervals. (66 is the number of intervals with 30 or more cars — the complement.)
Answer: A) 34
The number of cars passing a crossing was recorded in 100 half-hour intervals. Class 0–10: 7, 10–20: 14, 20–30: 13, 30–40: 12, 40–50: 20, 50–60: 11, 60–70: 15, 70–80: 8. In how many intervals did fewer than 30 cars pass?
Class 10StatisticsStatistics and Probability
Key idea: Median for grouped data
**Step 1**: build cumulative frequency (cf) column.
**Step 2**: find n/2 (half the total frequency).
**Step 3**: locate the **median class** — the FIRST class whose cumulative frequency ≥ n/2.
**Step 4**: Apply formula.
**Median formula**:
**Median = l + h × (n/2 − cf) / f**
Where:
- l = lower limit of median class
- h = class width
- n = total frequency
- cf = cumulative frequency of class BEFORE median class
- f = frequency of median class
The formula uses linear interpolation within the median class to find where the n/2-th item falls.
**Step 2**: find n/2 (half the total frequency).
**Step 3**: locate the **median class** — the FIRST class whose cumulative frequency ≥ n/2.
**Step 4**: Apply formula.
**Median formula**:
**Median = l + h × (n/2 − cf) / f**
Where:
- l = lower limit of median class
- h = class width
- n = total frequency
- cf = cumulative frequency of class BEFORE median class
- f = frequency of median class
The formula uses linear interpolation within the median class to find where the n/2-th item falls.
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