UP Board Solutions Class 9 Maths Chapter 20 Statistics Ex 20.5

Balaji Class 9 Maths Solutions Chapter 20 Statistics Ex 20.5 सांख्यिकी

Question 1. निम्नलिखित आँकड़ों का माध्य ज्ञात कीजिए :

Answer:
हलः

समान्तर माध्य \( \bar{x} = \frac{\Sigma fx}{N} = \frac{2650}{106} = 25 \)
In simple words: To find the mean of this grouped data, we multiply each 'x' value by its corresponding frequency 'f', sum these products (Σfx), and then divide by the total number of observations (N).

🎯 Exam Tip: Always double-check your calculations for Σfx and N to avoid errors in the final mean value. Neatly organized tables help prevent mistakes.

 

Question 2. निम्नलिखित बंटन के लिए माध्य की गणना कीजिए :

Answer:
हलः

समान्तर माध्य \( \bar{x} = \frac{\Sigma fx}{N} = \frac{557}{88} = 6.33 \)
In simple words: The mean for this distribution is calculated by summing the products of each value (x) and its frequency (f), then dividing by the total frequency. This gives the average value of the dataset.

🎯 Exam Tip: When dealing with decimal answers, ensure you round correctly to the specified number of decimal places, usually two or three, unless otherwise stated.

 

Question 3. निम्नलिखित के माध्य की गणना कीजिए :

Answer:
हलः

समान्तर माध्य \( \bar{x} = \frac{\Sigma fx}{N} = \frac{106000}{350} = 302.86 \)
In simple words: The mean is found by calculating the total sum of all values (Σfx) and dividing it by the total count of observations (N). This represents the average value for the given data distribution.

🎯 Exam Tip: Be careful with large numbers and multiple zeros in calculations. Use a calculator carefully for multiplication and summation to avoid arithmetic errors.

 

Question 4. एक फैक्ट्री के 60 श्रमिकों की सैलरी नीचे दी गई है।

Answer:
हलः

माध्य सैलरी \( = \frac{\Sigma fx}{N} = \frac{152500}{60} = 2541.70 \)
In simple words: The mean salary is found by calculating the total sum of salaries (Σfx) for all workers and then dividing by the total number of workers (N). This gives the average salary paid in the factory.

🎯 Exam Tip: When the total frequency (N) is given, ensure your sum of 'f' column matches this given value, as it can be a quick check for correctness.

 

Question 5. निम्नलिखित बंटन का माध्य ज्ञात कीजिए :

Answer:
हलः

समान्तर माध्य \( \bar{x} = \frac{\Sigma fx}{N} = \frac{28788}{250} = 115.15 \)
In simple words: The mean length is calculated by summing the products of each length value and its corresponding frequency, then dividing this sum by the total number of children. This gives the average length for the observed group.

🎯 Exam Tip: When working with grouped data, ensure the values of 'x' and 'f' are correctly multiplied. A miscalculation in any 'fx' product will lead to an incorrect sum and ultimately a wrong mean.

 

Question 6. यदि निम्नलिखित बंटन का माध्य 20 है। तो P का अज्ञात मान ज्ञात कीजिए।

Answer:
हलः

समान्तर माध्य \( \bar{x} = \frac{\Sigma fx}{N} \)

\( 20 = \frac{295 + 100p + 5p^2}{15 + 5p} \)

\( 20(15 + 5p) = 295 + 100p + 5p^2 \)

\( 300 + 100p = 295 + 100p + 5p^2 \)

\( 5p^2 = 300 - 295 \)

\( 5p^2 = 5 \)

\( p^2 = \frac{5}{5} \)

\( p^2 = 1 \)

\( p = \sqrt{1} = 1 \)
In simple words: To find the unknown value 'P', we set up the mean formula using the given data, which includes 'P'. We then solve the resulting algebraic equation, which simplifies to a quadratic equation, to find the value of 'P'.

🎯 Exam Tip: Pay close attention to algebraic manipulations when solving for an unknown variable. Distribute terms correctly and simplify carefully to arrive at the right answer.

 

Question 7. पौधों की माध्य लम्बाई ज्ञात कीजिए।
Answer:
हलः

पौधो की लम्बाई का समान्तर माध्य या माध्य लम्बाई \( = \frac{\Sigma fx}{N} = \frac{3459}{50} = 69.18 \)
In simple words: The mean length of the plants is calculated by summing the products of each plant length and its frequency, then dividing this total by the sum of all frequencies (total number of plants). This gives the average length.

🎯 Exam Tip: When calculating the mean, it's helpful to organize your data in a frequency table with an 'fx' column to ensure accurate multiplication and summation.

 

Question 8. निम्नलिखित बंटन के लिए x का अज्ञात मान ज्ञात कीजिए जिसका माध्य 31.87 है:

Answer:
हलः

माध्य \( \bar{x} = \frac{\Sigma fx}{N} \)

\( 31.87 = \frac{5114 + 30x}{200} \)

\( 200 \times 31.87 = 5114 + 30x \)

\( 6374 = 5114 + 30x \)

\( 6374 - 5114 = 30x \)

\( 1260 = 30x \)

\( x = \frac{1260}{30} \)

\( x = 42 \)
In simple words: We find the unknown value 'x' by setting up the mean formula using the given mean and the sums of 'f' and 'fx'. The equation is then solved algebraically to isolate and calculate 'x'.

🎯 Exam Tip: When the mean is given and an 'x' value is unknown, form an equation using the mean formula. Be careful with calculations involving decimals and algebraic expressions.

 

Question 9. a का मान ज्ञात कीजिए, यदि निम्नलिखित का माध्य 15 है:
Answer:
हलः

समान्तर माध्य \( \bar{x} = \frac{\Sigma fx}{N} \)

\( 15 = \frac{445 + 10a}{27 + a} \)

\( 15(27 + a) = 445 + 10a \)

\( 405 + 15a = 445 + 10a \)

\( 15a - 10a = 445 - 405 \)

\( 5a = 40 \)

\( a = \frac{40}{5} \)

\( a = 8 \)
In simple words: To find the unknown frequency 'a', we use the given mean and the expressions for total frequency (N) and sum of products (Σfx). Solving the resulting linear equation for 'a' provides the answer.

🎯 Exam Tip: When an unknown 'a' is in the frequency, remember it affects both N and Σfx. Set up the equation for the mean carefully and solve for 'a' using basic algebra.

 

Question 10. निम्नलिखित बंटन में अज्ञात बारंबारतायें ज्ञात कीजिए, यदि यह ज्ञात है कि बंटन का माध्य 50 है:

Answer:
हलः

समान्तर माध्य \( \bar{x} = \frac{\Sigma fx}{N} \)

\( 50 = \frac{3480 + 30f_1 + 70f_2}{68 + f_1 + f_2} \)

\( 50(68 + f_1 + f_2) = 3480 + 30f_1 + 70f_2 \)

\( 3400 + 50f_1 + 50f_2 = 3480 + 30f_1 + 70f_2 \)

\( 20f_1 - 20f_2 = 80 \)

\( f_1 - f_2 = 4 \) ...(1)
या
तथा \( 68 + f_1 + f_2 = 120 \) (दिया है)

\( f_1 + f_2 = 120 - 68 \)

\( f_1 + f_2 = 52 \) ...(2)
जोड़ने पर, \( (f_1 - f_2) + (f_1 + f_2) = 4 + 52 \)

\( 2f_1 = 56 \)

\( f_1 = \frac{56}{2} \)

\( f_1 = 28 \)
समीकरण (1) में \( f_1 \) का मान रखने पर \( 28 - f_2 = 4 \)

\( -f_2 = 4 - 28 \)

\( -f_2 = -24 \)

\( f_2 = 24 \)
In simple words: To find the unknown frequencies, we use the given mean and total frequency to form two simultaneous linear equations involving the unknown frequencies. Solving these equations gives us the values of \( f_1 \) and \( f_2 \).

🎯 Exam Tip: For problems with two unknown frequencies, always set up two equations: one from the mean formula and another from the total frequency. Solve these simultaneous equations carefully.