Statisticsसांख्यिकी

Bar heights: category 0 = 8, 1 = 15, 2 = 22, 3 = 30, 4 = 17, 5 = 8. Highest frequency is category 3 (30). Lowest non-zero frequency is 8, shared by categories 0 and 5. Differences: |3 - 0| = 3 and |3 - 5| = 2, so the minimum is 2,. Option D.

Read the frequency of each class from the polygon: 0-5 is 5, 5-10 is 12, 10-15 is 20, 15-20 is 18, 20-25 is 10 and 25-30 is 5. Total = 5 + 12 + 20 + 18 + 10 + 5 = 70. Answer: option B. Option B.

Frequencies: 30-40: 0, 40-50: 5, 50-60: 12, 60-70: 25, 70-80: 30, 80-90: 18, 90-100: 10. Students below the pass mark of 50 are in the classes 30-40 and 40-50, so failures = 0 + 5 = 5,. Option B.

In a histogram the area of each bar is proportional to the frequency, and the height is the frequency density (frequency divided by class width). Option B is correct. Option A is the width of the bar, and option C is the midpoint of a class, which is not drawn as a height. Option B.

From the bar graph, errors 0, 1, 2, 3, 4, 5 have 8, 15, 22, 30, 17, 8 participants, a total of $N=100$. The sum of $x\times f$ is $0+15+44+90+68+40=257$, so the mean is $2.57$ and $a=3$. The closed interval is $[2,4]$, so the values outside it are 0, 1 and 5, giving $8+15+8=31$, which is. Option A.

From the bar graph, the frequencies are: error 0: 8, error 1: 15, error 2: 22, error 3: 30, error 4: 17, error 5: 8. The total is $N=100$. Mean $=\frac{0\cdot8+1\cdot15+2\cdot22+3\cdot30+4\cdot17+5\cdot8}{100}=\frac{257}{100}=2.57$. Cumulative frequencies are 8, 23, 45, 75, so the 50th and 51st observations are both 3, and the median is 3. The absolute difference is $|3-2.57|=0.43$. Option B is correct. Option B.

Frequencies: 55, 80, 108, 117, 150, 105, 70. Total N = 685, so N/2 = 342.5. Cumulative frequencies are 55, 135, 243, 360, ... The first cumulative frequency at or above 342.5 is 360, which belongs to the 180-190 class. Answer: D. Option D.

Weekday frequencies from the polygon are 0, 34, 100, 85, 30, 22, 0, total 271, so N/2 = 135.5. Cumulative frequency up to 60-90 is 134, so the median class is 90-120 with f = 85, F = 134 and h = 30. Median = 90 + ((135.5 - 134) / 85) x 30 = 90 + 0.53 = 90.5 minutes,. Option B.

The frequency polygon peaks at 18 students for the class 45-60, which is the highest frequency. The midpoint of a class is $\frac{\text{lower limit}+\text{upper limit}}{2}=\frac{45+60}{2}=52.5$. So the answer is option A. Option B (52) is only a rounded value, while C (45.2) and D (60.5) are not the midpoint of any class here. Option A.

Frequencies by attempts: 1 → 14, 2 → 26, 3 → 35, 4 → 15, 5 → 7, 6 → 3. Total N = 100. Cumulative frequencies: 14, 40, 75, 90, 97, 100. The 50th and 51st values both fall in attempt 3, so the median is 3. The mode is also 3 (frequency 35). The absolute difference is 0. Answer: A. Option A.

Frequencies are 55, 80, 108, 117, 150, 105 and 70, so the total is 685. The modal class is 190-200 (highest frequency 150). Probability = 150 / 685 = 0.219, which rounds to 0.22. Answer:. Option B.

HR is 10% and Operations is 15%, so together they are 25% of employees. Probability = 0.25. Answer:. Option C.

Frequencies from the histogram: 40-45 = 5, 45-50 = 8, 50-55 = 12, 55-60 = 10, 60-65 = 7, 65-70 = 3, total 45. One-fourth of 45 is 11.25, and the class 50-55 with 12 people is the closest. Option B is correct. Option B.