Mean = sum of the numbers/total numbers

Mean = \frac{35 + 17 + 22 + 26+ 29 + 34 + x + 42 + 67 + 70 + y}{11}

So 42 = \frac{342 + x + y}{11}

\Rightarrow x + y = 462 - 342 = 120. Assume this to be equation 1.

The number of terms is 10, so the median will be the average of the n^{th} and \frac{n+1}{2}^{th} term or the 5^{th} and the 6^{th} term.

Median = 35, so \frac{34 + x}{2} = 35

Solve for x to get x = 36

Plug this back in equation 1 and solve for y to get y = 120 - 36 = 84

So, x and y are 36, 84.

I found an answer from www.analyzemath.com

**Statistics** and **Probability Problems** with **Answers**

**Find x** and **y so** that the **ordered data set has** a **mean** of **42** and a median of 35. 17 , 22 , 26 , 29 , 34 , **x** , **42** , 67 , 70 , **y**. Given ...

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