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
Let the Mean and Variance of five observations $ x_i $, $ i = 1, 2, 3, 4, 5 $ be 5 and 10 respectively. If three observations are $ x_1 = 1, x_2 = 3, x_3 = a $ and $ x_4 = 7, x_5 = b $ with $ a>b $, then the Variance of the observations $ n + x_n $ for $ n = 1, 2, 3, 4, 5 $ is
Find the mean of the following distribution:
\[\begin{array}{|c|c|c|c|c|c|c|c|} \hline \textbf{Class-interval} & 11-13 & 13-15 & 15-17 & 17-19 & 19-21 & 21-23 & 23-25 \\ \hline \text{Frequency} & \text{7} & \text{6} & \text{9} & \text{13} & \text{20} & \text{5} & \text{4} \\ \hline \end{array}\]
‘दीवार खड़ी करना’ मुहावरे का वाक्य में इस प्रकार प्रयोग करें कि अर्थ स्पष्ट हो जाए।
Select from the following a statement which is not true about the burning of magnesium ribbon in air:
Analyze the significant changes in printing technology during 19th century in the world.
निम्नलिखित विषय पर संकेत बिंदुओं के आधार पर लगभग 120 शब्दों में एक अनुच्छेद लिखिए |
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