Answer

(a) $E[X] = \sum_{x=0}^{3} x P(X = x) = 0(0.3) + 1(0.4) + 2(0.2) + 3(0.1) = 0 + 0.4 + 0.4 + 0.3 = 1.1$

(b) $Var(X) = \sum_{x=0}^{3} (x - 1.1)^2 P(X = x) = (0 - 1.1)^2(0.3) + (1 - 1.1)^2(0.4)$

$+ (2 - 1.1)^2(0.2) + (3 - 1.1)^2(0.1) = (1.21)(0.3) + (0.01)(0.4) + (0.81)(0.2) + (3.61)(0.1)$ ,

$= 0.363 + 0.004 + 0.162 + 0.361 = 0.89$

(c) Our hypotheses are $H_0: \mu = 1.1$ vs $H_1: \mu > 1.1$.

(d) For a $5\%$ one-tailed (upper tail) test, the critical value is $1.645$. The test statistic is calculated as: \[ z = \frac{1.28 - 1.1}{\sqrt{0.89 / 100}} = \frac{0.18}{0.09434} \approx 1.91 \]

(e) Since $1.91 > 1.645$, we reject the null hypothesis of accurate ticket recording. This suggests there is a statistically significant inflation in reported ticket complexity, supporting the suspicion of padding.

Answer

(a) $E[X] = \sum_{x=1}^{4} x P(X = x) = 1(0.2) + 2(0.5) + 3(0.2) + 4(0.1) = 0.2 + 1.0 + 0.6 + 0.4 = 2.2$

(b) $Var(X) = \sum_{x=1}^{4} (x - 2.2)^2 P(X = x) = (1 - 2.2)^2(0.2) + (2 - 2.2)^2(0.5) $

$ + (3 - 2.2)^2(0.2) + (4 - 2.2)^2(0.1) = (1.44)(0.2) + (0.04)(0.5) + (0.64)(0.2) + (3.24)(0.1) $

$= 0.288 + 0.02 + 0.128 + 0.324 = 0.76$

(c) Our hypotheses are $H_0: \mu = 2.2$ vs $H_1: \mu > 2.2$.

(d) For a $5\%$ one-tailed (upper tail) test, the critical value is $1.645$. The test statistic is: \[ z = \frac{2.42 - 2.2}{\sqrt{0.76 / 100}} = \frac{0.22}{0.08718} \approx 2.52 \]

(e) Since $2.52 > 1.645$, we reject the null hypothesis of honesty. There is clear evidence suggesting that drivers are over-utilizing the car wash allowance abnormally, pointing to likely corporate card abuse.

Answer

(a) Let $\bar{x}$ be the sample proportion of respondents who answered "yes". Mathematically, question (a) asks for $P[\bar{X} \le 0.32]$ when $p = 0.333$. So we have:

\begin{align*} P[\bar{X} \le 0.32] &= P[\bar{X} - 0.333 \le 0.32 - 0.333] \\ &= P\left[ \frac{\bar{X} - 0.333}{\sqrt{0.333 * 0.667 / 1500}} \le \frac{0.32 - 0.333}{\sqrt{0.333 * 0.667 / 1500}} \right] \\ &\approx P[Z \le -1.07] \\ &= 14.23\% \end{align*}

where the approximation is the use of the central limit theorem. This is justified here by assuming that the organization used a randomized strategy to ensure independence.

(b) Certainly not too unreasonable, since it happens with about $14.2\%$ by chance that we would get a sample proportion of $32\%$ or less. So yes, it is reasonable.

(c) The t-test gives a value of $-1.07$, and the critical value is $-1.645$, so no we would not reject. Notice that the answer in (b) and (c) are very related, once you have done it one way you have the basic calculations to do it the other way (hypothesis test vs p-value).

Answer

(a) Let $\bar{x}$ be the sample proportion of questions answered "yes". In math, this question asks for $P[\bar{X} \le 0.415]$ when $p = 0.43$. So we have:

\begin{align*} P[\bar{X} \le 0.415] &= P[\bar{X} - 0.43 \le 0.415 - 0.43] \\ &= P\left[ \frac{\bar{X} - 0.43}{\sqrt{0.43 * 0.57 / 2000}} \le \frac{0.415 - 0.43}{\sqrt{0.43 * 0.57 / 2000}} \right] \\ &\approx P[Z \le -1.35] \\ &= 8.85\% \end{align*}

where the normal approximation is applied via the central limit theorem under independent random selection.

(b) Yes, this is completely reasonable. There is an $8.85\%$ chance that standard sampling fluctuations would yield a sample value of $41.5\%$ or lower if the true baseline population parameter were $43\%$.

(c) We are conducting a 5% one tail test, and in (b) show that there is an $8.85\%$ of seeing a lower number. So we do not reject.

Answer

(a) We want $P(\bar{X} \ge 65)$. Given $\mu = 60$, $\sigma = 20$, and $n = 45$, we have:

\begin{align*} P(\bar{X} \ge 65) &= P(\bar{X} - 60 \ge 65 - 60) \\ &= P\left( \frac{\bar{X} - 60}{20/\sqrt{45}} \ge \frac{65 - 60}{20/\sqrt{45}} \right) \\ &\approx P(Z \ge 1.6771) \\ &= 0.0468 \end{align*}

There is roughly a $4.68\%$ chance of seeing a sample mean this large by random variation.

(b) Since the probability ($4.68\%$) falls just below the typical $5\%$ threshold, it is statistically significant but borderline. It suggests that this group might be exercising slightly longer than standard historical baselines, though random noise remains a minor possibility.

(c) The Central Limit Theorem is valid because $n = 45 > 30$. Gym session lengths are naturally bounded on the lower side (rarely less than 15-20 minutes) but can have right-skewness from extreme power-lifters or endurance athletes. The sample size is large enough to normalize this skewness. Independence holds assuming the selected members were sampled randomly across different times of day rather than all belonging to a single specialized class.

(d) We wish to compute $P(58 < \bar{X} < 62)$: \begin{align*} P(58 < \bar{X} < 62) &= P(58 - 60 < \bar{X} - 60 < 62 - 60) \\ &= P\left( \frac{58 - 60}{20/\sqrt{45}} < \frac{\bar{X} - 60}{20/\sqrt{45}} < \frac{62 - 60}{20/\sqrt{45}} \right) \\ &\approx P(-0.6708 < Z < 0.6708) \\ &= 0.4976 \end{align*} We will see sample averages fall within this close 4-minute window approximately half ($49.8\%$) of the time.

Answer

(a) We want $P(\bar{X} \ge 510)$. Given $\mu = 400$, $\sigma = 300$, and $n = 30$, we have: \begin{align*} P(\bar{X} \ge 510) &= P(\bar{X} - 400 \ge 510 - 400) \\ &= P\left( \frac{\bar{X} - 400}{300/\sqrt{30}} \ge \frac{510 - 400}{300/\sqrt{30}} \right) \\ &\approx P(Z \ge 2.0083) \\ &= 0.0223 \end{align*} There is around a $2.23\%$ chance of seeing a monthly average this high or higher by pure random variation.

(b) Just through randomness of sales, we would not have expected to obtain such a large monthly average, since a probability of $2.23\%$ is low. It seems as though sales are genuinely higher than historically expected (the real mean is likely higher than \$400).

(c) We are relying on the Central Limit Theorem. However, daily revenue for an e-commerce site is highly likely to be strongly right-skewed (many small days, a few huge checkout days), and $n=30$ might be a bit small to completely smooth out extreme skewness. Furthermore, consecutive days might violate independence due to weekly patterns (weekends vs. weekdays) or multi-day marketing promotions.

(d) Here we wish to compute $P(380 < \bar{X} < 420)$: \begin{align*} P(380 < \bar{X} < 420) &= P(380 - 400 < \bar{X} - 400 < 420 - 400) \\ &= P\left( \frac{380 - 400}{300/\sqrt{30}} < \frac{\bar{X} - 400}{300/\sqrt{30}} < \frac{420 - 400}{300/\sqrt{30}} \right) \\ &\approx P(-0.3651 < Z < 0.3651) \\ &= 0.2848 \end{align*} We will see such sample means from this problem roughly $28.5\%$ of the time.