Class : | 4 – 6 | 7 – 9 | 10 – 12 | 13 – 15 |
Frequency : | 5 | 4 | 9 | 10 |
Step 1: Find the class midpoints
The midpoints ($x_i$) are calculated as the average of the lower and upper limits of each class:
\[x_1 = \frac{4 + 6}{2} = 5, \quad x_2 = \frac{7 + 9}{2} = 8, \quad x_3 = \frac{10 + 12}{2} = 11, \quad x_4 = \frac{13 + 15}{2} = 14.\]
Step 2: Multiply midpoints by frequencies
Construct the table as follows:
\[\begin{array}{|c|c|c|c|}\hline\text{Class} & \text{Frequency } (f_i) & \text{Midpoint } (x_i) & f_i x_i \\\hline4 - 6 & 5 & 5 & 25 \\7 - 9 & 4 & 8 & 32 \\10 - 12 & 9 & 11 & 99 \\13 - 15 & 10 & 14 & 140 \\\hline\end{array}\]
Step 3: Calculate the total frequency and total $f_i x_i$
\[\sum f_i = 5 + 4 + 9 + 10 = 28.\]
\[\sum f_i x_i = 25 + 32 + 99 + 140 = 296.\]
Step 4: Calculate the mean
The formula for the mean is:
\[\text{Mean} = \frac{\sum f_i x_i}{\sum f_i}.\]
Substitute the values:
\[\text{Mean} = \frac{296}{28} = 10.57.\]
Correct Answer: Mean = 10.57
Class | 0 – 15 | 15 – 30 | 30 – 45 | 45 – 60 | 60 – 75 | 75 – 90 |
---|---|---|---|---|---|---|
Frequency | 11 | 8 | 15 | 7 | 10 | 9 |
Variance of the following discrete frequency distribution is:
\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Class Interval} & 0-2 & 2-4 & 4-6 & 6-8 & 8-10 \\ \hline \text{Frequency (}f_i\text{)} & 2 & 3 & 5 & 3 & 2 \\ \hline \end{array} \]
"जितेंद्र नार्गे जैसे गाइड के साथ किसी भी पर्यटन स्थल का भ्रमण अधिक आनंददायक और यादगार हो सकता है।" इस कथन के समर्थन में 'साना साना हाथ जोड़ि .......' पाठ के आधार पर तर्कसंगत उत्तर दीजिए।
आप अदिति / आदित्य हैं। आपकी दादीजी को खेलों में अत्यधिक रुचि है। ओलंपिक खेल-2024 में भारत के प्रदर्शन के बारे में जानकारी देते हुए लगभग 100 शब्दों में पत्र लिखिए।
There is a circular park of diameter 65 m as shown in the following figure, where AB is a diameter. An entry gate is to be constructed at a point P on the boundary of the park such that distance of P from A is 35 m more than the distance of P from B. Find distance of point P from A and B respectively.