The mean of the ages of five friends is 16 years. If three of them are of the same age and theother two are of 14 and 18 years, then the age of each friends having the same age isA. 12 years.B. 16 years.C. 18 years.D. 20 years.9. For the data set 4, 0, 4 and 0, the variance isA. 0B. 1C. 2D. 4
Question
The mean of the ages of five friends is 16 years. If three of them are of the same age and theother two are of 14 and 18 years, then the age of each friends having the same age isA. 12 years.B. 16 years.C. 18 years.D. 20 years.9. For the data set 4, 0, 4 and 0, the variance isA. 0B. 1C. 2D. 4
Solution
Sure, let's solve these problems one by one.
- The mean of the ages of five friends is 16 years. If three of them are of the same age and the other two are of 14 and 18 years, then the age of each friend having the same age is:
First, we need to find the total sum of the ages. Since the mean is 16 and there are 5 friends, the total sum of their ages is 16 * 5 = 80 years.
We know that two of the friends are 14 and 18 years old. So, the combined age of these two friends is 14 + 18 = 32 years.
The remaining three friends must therefore have a combined age of 80 - 32 = 48 years.
Since these three friends are all the same age, we can find their individual age by dividing the total by 3. So, 48 / 3 = 16 years.
So, the answer is B. 16 years.
- For the data set 4, 0, 4 and 0, the variance is:
First, we need to find the mean of the data set. The mean is (4 + 0 + 4 + 0) / 4 = 2.
Next, we subtract the mean from each data point and square the result: (4-2)^2 = 4, (0-2)^2 = 4, (4-2)^2 = 4, (0-2)^2 = 4.
Then, we find the mean of these squared differences: (4 + 4 + 4 + 4) / 4 = 4.
So, the variance is D. 4.
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