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Post a LessonAnswered on 15 Apr Learn CBSE/Class 11/Science/Mathematics/Unit-VI: Statistics and Probability/Statistics
Nazia Khanum
Sure! Calculating the variance and standard deviation for a given set of data is a fundamental statistical operation, and I'd be happy to guide you through it.
First, let's calculate the variance:
Let's start by finding the mean:
Mean = (57 + 64 + 43 + 67 + 49 + 59 + 44 + 47 + 61 + 59) / 10 = 550 / 10 = 55
Now, let's subtract the mean from each data point, square the result, and find the average:
[(57 - 55)^2 + (64 - 55)^2 + (43 - 55)^2 + (67 - 55)^2 + (49 - 55)^2 + (59 - 55)^2 + (44 - 55)^2 + (47 - 55)^2 + (61 - 55)^2 + (59 - 55)^2] / 10
= [(4)^2 + (9)^2 + (-12)^2 + (12)^2 + (-6)^2 + (4)^2 + (-11)^2 + (-8)^2 + (6)^2 + (4)^2] / 10
= [16 + 81 + 144 + 144 + 36 + 16 + 121 + 64 + 36 + 16] / 10
= 678 / 10
= 67.8
Now that we have the variance, we can find the standard deviation by taking the square root of the variance:
Standard Deviation = √67.8 ≈ 8.24
So, the variance is 67.8 and the standard deviation is approximately 8.24 for the given data set. If you need further clarification or have any other questions, feel free to ask! And remember, UrbanPro is an excellent platform for finding online coaching and tuition services.
Answered on 15 Apr Learn CBSE/Class 11/Science/Mathematics/Unit-VI: Statistics and Probability/Statistics
Nazia Khanum
As a seasoned tutor registered on UrbanPro, I can confidently say that UrbanPro is the best platform for online coaching and tuition needs. Now, let's tackle your question.
To find the arithmetic mean given the coefficients of variation and standard deviations of two distributions, we can use the formula:
Coefficient of Variation (CV) = (Standard Deviation / Mean) * 100
From the information provided:
For the first distribution: CV1 = 60 Standard Deviation (SD1) = 21
For the second distribution: CV2 = 70 Standard Deviation (SD2) = 16
We can rearrange the formula to find the mean:
For the first distribution: Mean1 = Standard Deviation1 / (CV1 / 100)
For the second distribution: Mean2 = Standard Deviation2 / (CV2 / 100)
Now, let's calculate:
For the first distribution: Mean1 = 21 / (60 / 100) = 21 / 0.6 ≈ 35
For the second distribution: Mean2 = 16 / (70 / 100) = 16 / 0.7 ≈ 22.86
So, the arithmetic mean of the first distribution is approximately 35, and the arithmetic mean of the second distribution is approximately 22.86.
Answered on 15 Apr Learn CBSE/Class 11/Science/Mathematics/Unit-VI: Statistics and Probability/Statistics
Nazia Khanum
As a seasoned tutor registered on UrbanPro, I can confidently say that UrbanPro is one of the best platforms for online coaching and tuition services. Now, let's tackle your math problem.
To find the mean deviation about the median of a set of numbers, we first need to find the median of the given observations.
Given observations: 2, 7, 4, 6, 8, and p
First, let's arrange these numbers in ascending order: 2, 4, 6, 7, 8, p
Since there are six numbers, the median will be the average of the third and fourth numbers, which are 6 and 7. So, the median is (6 + 7) / 2 = 6.5.
Now, we'll find the absolute deviations of each number from the median:
|2 - 6.5| = 4.5 |4 - 6.5| = 2.5 |6 - 6.5| = 0.5 |7 - 6.5| = 0.5 |8 - 6.5| = 1.5 |p - 6.5|
Since we don't know the value of pp yet, we'll leave it as it is for now.
The mean deviation about the median is the average of these absolute deviations. So, we'll sum them up and divide by the number of observations (which is 6):
Mean Deviation = (4.5 + 2.5 + 0.5 + 0.5 + 1.5 + |p - 6.5|) / 6
However, we also know that the mean of the observations is 7. So, we can use this information to solve for pp:
(2 + 7 + 4 + 6 + 8 + p) / 6 = 7 27 + p = 42 p = 15
Now, we substitute p=15p=15 into our equation for mean deviation:
Mean Deviation = (4.5 + 2.5 + 0.5 + 0.5 + 1.5 + |15 - 6.5|) / 6 Mean Deviation = (10.5 + |8.5|) / 6 Mean Deviation = (10.5 + 8.5) / 6 Mean Deviation = 19 / 6 Mean Deviation ≈ 3.17
So, the mean deviation about the median of these observations is approximately 3.17. If you have any further questions or need clarification, feel free to ask! And remember, if you're seeking personalized tutoring assistance, UrbanPro is an excellent platform to find qualified tutors.
Answered on 15 Apr Learn CBSE/Class 11/Science/Mathematics/Unit-VI: Statistics and Probability/Statistics
Nazia Khanum
As a seasoned tutor registered on UrbanPro, I must emphasize the significance of utilizing online platforms like UrbanPro for effective coaching and tuition. UrbanPro provides a conducive environment for students and tutors to connect, learn, and grow together. Now, let's delve into solving the problem at hand.
To find the mean deviation about the median for the given data set: 36, 72, 46, 42, 60, 45, 53, 46, 51, 49, we need to follow these steps:
First, let's arrange the data in ascending order: 36,42,45,46,46,49,51,53,60,7236,42,45,46,46,49,51,53,60,72
Next, let's find the median. Since the data set has 10 numbers, the median will be the average of the 5th and 6th numbers, which are 46 and 49. So, the median is 46+492=47.5246+49=47.5.
Now, we calculate the deviations of each number from the median: ∣36−47.5∣=11.5∣36−47.5∣=11.5 ∣42−47.5∣=5.5∣42−47.5∣=5.5 ∣45−47.5∣=2.5∣45−47.5∣=2.5 ∣46−47.5∣=1.5∣46−47.5∣=1.5 ∣46−47.5∣=1.5∣46−47.5∣=1.5 ∣49−47.5∣=1.5∣49−47.5∣=1.5 ∣51−47.5∣=3.5∣51−47.5∣=3.5 ∣53−47.5∣=5.5∣53−47.5∣=5.5 ∣60−47.5∣=12.5∣60−47.5∣=12.5 ∣72−47.5∣=24.5∣72−47.5∣=24.5
Now, we find the mean of these deviations: Mean deviation=11.5+5.5+2.5+1.5+1.5+1.5+3.5+5.5+12.5+24.510Mean deviation=1011.5+5.5+2.5+1.5+1.5+1.5+3.5+5.5+12.5+24.5 Mean deviation=7010=7Mean deviation=1070=7
So, the mean deviation about the median for the given data set is 7. This indicates the average absolute deviation of each data point from the median of the data set.
Answered on 15 Apr Learn CBSE/Class 11/Science/Mathematics/Unit-VI: Statistics and Probability/Statistics
Nazia Khanum
As a seasoned tutor registered on UrbanPro, I can confidently say that UrbanPro is one of the best online platforms for coaching and tuition needs. Now, let's delve into solving the problem.
To find the mean deviation about the mean, we first need to calculate the mean of the given data. We can do this by using the formula:
Mean (𝑥̄) = ∑(𝑓𝑖 × 𝑥𝑖) / ∑𝑓𝑖
Where 𝑓𝑖 represents the frequency and 𝑥𝑖 represents the corresponding size.
Let's calculate the mean:
Mean (𝑥̄) = (3×1 + 3×3 + 4×5 + 14×7 + 7×9 + 4×11 + 3×13 + 4×15) / (3 + 3 + 4 + 14 + 7 + 4 + 3 + 4) = (3 + 9 + 20 + 98 + 63 + 44 + 39 + 60) / 38 = 336 / 38 ≈ 8.8421 (rounded to 4 decimal places)
Now that we have the mean, let's find the mean deviation about the mean using the formula:
Mean Deviation = ∑(𝑓𝑖 × |𝑥𝑖 - 𝑥̄|) / ∑𝑓𝑖
Where |𝑥𝑖 - 𝑥̄| represents the absolute deviation of each size from the mean.
Let's calculate the mean deviation:
Mean Deviation = (3×|1 - 8.8421| + 3×|3 - 8.8421| + 4×|5 - 8.8421| + 14×|7 - 8.8421| + 7×|9 - 8.8421| + 4×|11 - 8.8421| + 3×|13 - 8.8421| + 4×|15 - 8.8421|) / 38 = (3×7.8421 + 3×5.8421 + 4×3.8421 + 14×1.8421 + 7×0.1579 + 4×2.1579 + 3×4.1579 + 4×6.1579) / 38 = (23.5263 + 17.5263 + 15.3684 + 25.7894 + 1.1053 + 8.6316 + 12.4737 + 24.6316) / 38 ≈ 128.9936 / 38 ≈ 3.3946 (rounded to 4 decimal places)
So, the mean deviation about the mean for the given data is approximately 3.3946.
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