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Worked solution

Translate between a sample mean and a total

The mean time spent by 12-year-old children watching television in one day is 3.5 hours, with standard deviation 0.7 hours.

(i) For a random sample of 100 children, find the probability that the mean watching time is between 3.4 and 3.7 hours.

(ii) Find the probability that their total Sunday watching time exceeds 365 hours.

CJC · Discussion · Question 7

Answer

(i) Approximately 0.9210.921; (ii) approximately 0.01610.0161.

(i) Apply CLT to the sample mean

Let independent XjX_j denote daily watching time, with mean 3.5 and variance 0.720.7^2. For n=100n=100, Xˉ∼˙N(3.5,0.72/100)=N(3.5,0.0049)\bar X\mathrel{\dot\sim}N(3.5,0.7^2/100)=N(3.5,0.0049) by the CLT. Thus P(3.4<Xˉ<3.7)≈0.921P(3.4<\bar X<3.7)\approx0.921, to 3 s.f.

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(ii) Method 1: express the total event as a mean event

For S=∑j=1100XjS=\sum_{j=1}^{100}X_j, S>365S>365 is equivalent to Xˉ>3.65\bar X>3.65. Therefore P(S>365)≈1−Φ[(3.65−3.5)/0.07]=0.0161P(S>365)\approx1-\Phi[(3.65-3.5)/0.07]=0.0161, to 3 s.f.

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(ii) Method 2: approximate the total directly

The total has mean 100(3.5)=350100(3.5)=350 and variance 100(0.72)=49100(0.7^2)=49. The CLT gives S∼˙N(350,49)S\mathrel{\dot\sim}N(350,49), hence P(S>365)≈1−Φ[(365−350)/7]=0.0161P(S>365)\approx1-\Phi[(365-350)/7]=0.0161. Both methods use the same event. Applying the given daily parameters to Sunday is an assumption implicit in the source model.

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Common mistakes

The mean has standard deviation 0.07 hours, while the total has standard deviation 7 hours. Keep the model’s independence and Sunday assumptions explicit.

Solution notes

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