Exercise 3 · Binomial, normal and confidence interval — Web loading time

Binomial distribution, normal approximation and a confidence interval for the mean.

Maximum score · 2.5 points

A telecommunications company considers a web page efficient if its loading time is under 3 seconds. The company claims that at least 50 % of the pages it manages are efficient. To check this claim, a random sample of $n = 2500$ web pages managed by the company is selected.

Formulas provided: $Z \sim N(0,1) \Rightarrow P(-1.96 \le Z \le 1.96) = 0.95$ and $P(-2.58 \le Z \le 2.58) = 0.99$; confidence intervals $\left[\hat{p} - z_{\gamma}\sqrt{\tfrac{\hat{p}(1-\hat{p})}{n}},\; \hat{p} + z_{\gamma}\sqrt{\tfrac{\hat{p}(1-\hat{p})}{n}}\right]$ for the proportion and $\left[\bar{x} - z_{\gamma}\tfrac{s}{\sqrt{n}},\; \bar{x} + z_{\gamma}\tfrac{s}{\sqrt{n}}\right]$ for the mean.

Assume that, each time a web page is selected for the sample, it can be efficient (or not) with probability $p = 0.5$, independently of all other pages. Consider the random variable $X$ counting how many of the 2500 sampled pages are efficient. What distribution does $X$ follow? Compute the probability that at most 1299 pages are efficient, using the normal approximation without continuity correction. 1.25 p
In the sample of $n = 2500$ web pages, the sample mean loading time was $\bar{x} = 2.95$ seconds with a sample standard deviation of $s = 0.38$ seconds. Build a 95 % confidence interval for the mean loading time of the company's pages. Based on the interval, what can be said about the company's claim that at least 50 % of its pages are efficient? 1.25 p

Bachillerato CCSS · Block E — Stochastic sense Binomial distribution Normal approximation Confidence interval

Step-by-step solution

Key idea: $X$ counts successes in $n$ independent trials with constant success probability: a binomial. Since $n$ is very large, it is approximated by a normal with mean $np$ and standard deviation $\sqrt{np(1-p)}$.

a) Distribution of $X$ and the probability

Each page is efficient with probability $p = 0.5$, independently, and we count successes among $n = 2500$:

X \sim B(2500;\; 0.5).

Parameters of the normal approximation:

\mu = np = 2500 \cdot 0.5 = 1250, \qquad \sigma = \sqrt{np(1-p)} = \sqrt{2500 \cdot 0.5 \cdot 0.5} = \sqrt{625} = 25.

Hence $X \approx N(1250,\, 25)$. Standardise (no continuity correction, as requested):

P(X \le 1299) = P\!\left(Z \le \frac{1299 - 1250}{25}\right) = P(Z \le 1.96).

Since $P(-1.96 \le Z \le 1.96) = 0.95$, each tail is $0.025$, so $P(Z \le 1.96) = 0.95 + 0.025 = 0.975$.

$X \sim B(2500;\, 0.5) \approx N(1250,\, 25)$ and $P(X \le 1299) = 0.975$.

b) Confidence interval and interpretation

Large sample with unknown variance: use the interval for the mean with $z_{\gamma} = 1.96$ (95 % confidence):

\bar{x} \pm z_{\gamma}\,\frac{s}{\sqrt{n}} = 2.95 \pm 1.96 \cdot \frac{0.38}{\sqrt{2500}} = 2.95 \pm 1.96 \cdot \frac{0.38}{50} = 2.95 \pm 0.0149.

CI_{95\%} = [\,2.9351,\; 2.9649\,]\ \text{seconds}.

The whole interval lies below 3 seconds: with 95 % confidence, the mean loading time of the company's pages is under 3 seconds.

This is consistent with — and supportive of — the company's claim: if the mean time is clearly below 3 s, it is plausible that at least half the pages load in under 3 s. (Strictly speaking, the interval concerns the mean while the claim concerns the proportion of efficient pages; the interval does not prove the claim, but it does not contradict it.)

$CI_{95\%} = [2.9351;\ 2.9649]$ s. Since the whole interval is below 3 s, the data support the company's claim.