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Surface Areas And Volumes
EXERCISE 13.1
CBSE_NCERT_TEXTBOOK
Grade 10
Question:
Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method. Number of heartbeats 65 - 68 68 - 71 71 - 74 74 - 77 77 - 80 80 - 83 83 - 86 per minute Number of women 2 4 3 8 7 4 2
Step-by-Step Solution
Key Concept: For grouped data, the mean is obtained by using the class‑mark (mid‑point) of each class as a representative value. The formula is \(\displaystyle \bar{x}=\frac{\sum f_i x_i}{\sum f_i}\), where \(f_i\) is the frequency of the i‑th class and \(x_i\) is its class‑mark.
1. Identify the class‑marks (mid‑points) for each class:\\ \[\begin{aligned} 65-68 &: \frac{65+68}{2}=66.5 \\ 68-71 &: \frac{68+71}{2}=69.5 \\ 71-74 &: \frac{71+74}{2}=72.5 \\ 74-77 &: \frac{74+77}{2}=75.5 \\ 77-80 &: \frac{77+80}{2}=78.5 \\ 80-83 &: \frac{80+83}{2}=81.5 \\ 83-86 &: \frac{83+86}{2}=84.5 \end{aligned}\] 2. Multiply each class‑mark by its frequency (\(f_i x_i\)) and tabulate:\\ \[\begin{array}{c|c|c} \text{Class} & f_i & f_i x_i \\ \hline 65-68 & 2 & 2\times66.5 = 133 \\ 68-71 & 4 & 4\times69.5 = 278 \\ 71-74 & 3 & 3\times72.5 = 217.5 \\ 74-77 & 8 & 8\times75.5 = 604 \\ 77-80 & 7 & 7\times78.5 = 549.5 \\ 80-83 & 4 & 4\times81.5 = 326 \\ 83-86 & 2 & 2\times84.5 = 169 \\ \hline \text{Total} & \sum f_i = 30 & \sum f_i x_i = 2277 \end{array}\] 3. Apply the mean formula:\\ \[\bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{2277}{30}=75.9\] 4. Interpretation: The average (mean) heart‑beat rate of the 30 women is 75.9 beats per minute.
Correct Answer:75.9 beats per minute
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