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Days for a Chebyshev guarantee

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Daily results are independent and identically distributed with variance $49$ . Using Chebyshev's inequality for the sample mean, what is the smallest whole number of days $n$ that guarantees $P (∣ X ˉ - μ ∣\geq 2) \leq 0.1$ ? Give a whole number.

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Hint

The bound is $\frac{σ ^{2}}{n ϵ ^{2}}$ : set it no higher than the target and solve for $n$ .

Solution

The bound is $\frac{49}{4 n}$ , so we need $n \geq \frac{49}{4 \times 0.1}$ , and $\frac{49}{4 \times 0.1} = 122.5$ .
Rounding up to a whole number of days gives $n = 123$ .

Answer: $123$

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