Question

a large citys transit department claims that only 10% of city buses run off - schedule. to test this claim, a random sample of 10 buses is chosen at random. five of the buses are running off - schedule. to see how unusual this sample of buses is, a simulation of 100 trials was conducted under the assumption that 10% of the buses run off - schedule. based on the dotplot of the simulation results and the sample of 10 buses, which conclusion can be drawn? it is most likely that exactly one out of 10 buses is running off - schedule. there is about a 3% chance that 5 or more buses are running off - schedule. this is unusual and is convincing evidence that the true probability that a bus is off - schedule is more than 10%.

Explanation:

Step1: Analyze the claim

The claim is that 10% of city - buses run off - schedule. In a sample of $n = 10$ buses, if the claim is true, the number of off - schedule buses $X$ follows a binomial distribution $X\sim B(n = 10,p=0.1)$.

Step2: Evaluate the first conclusion

The probability that exactly one out of 10 buses is running off - schedule in a binomial distribution is given by the formula $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n = 10$, $k = 1$, and $p = 0.1$. $C(10,1)=\frac{10!}{1!(10 - 1)!}=10$, $p = 0.1$, $1-p = 0.9$. So $P(X = 1)=10\times0.1\times0.9^{9}\approx0.387$. It is not "most likely" that exactly one out of 10 buses is running off - schedule.

Step3: Evaluate the second conclusion

We need to find $P(X\geq5)=\sum_{k = 5}^{10}C(10,k)\times0.1^{k}\times0.9^{10 - k}$.
$C(10,5)=\frac{10!}{5!(10 - 5)!}=252$, $C(10,6)=\frac{10!}{6!(10 - 6)!}=210$, $C(10,7)=\frac{10!}{7!(10 - 7)!}=120$, $C(10,8)=\frac{10!}{8!(10 - 8)!}=45$, $C(10,9)=\frac{10!}{9!(10 - 9)!}=10$, $C(10,10)=\frac{10!}{10!(10 - 10)!}=1$.
$P(X = 5)=252\times0.1^{5}\times0.9^{5}\approx0.00157$, $P(X = 6)=210\times0.1^{6}\times0.9^{4}\approx0.000138$, $P(X = 7)=120\times0.1^{7}\times0.9^{3}\approx0.0000088$, $P(X = 8)=45\times0.1^{8}\times0.9^{2}\approx0.00000041$, $P(X = 9)=10\times0.1^{9}\times0.9^{1}\approx0.000000009$, $P(X = 10)=1\times0.1^{10}\times0.9^{0}\approx0.0000000001$.
$P(X\geq5)=P(X = 5)+P(X = 6)+P(X = 7)+P(X = 8)+P(X = 9)+P(X = 10)\approx0.00157+0.000138 + 0.0000088+0.00000041+0.000000009+0.0000000001\approx0.00172\approx0.2\%$. But if we assume the simulation results are valid and say there is about a 3% chance that 5 or more buses are running off - schedule, this is an unusual event under the assumption that $p = 0.1$. So it is convincing evidence that the true probability that a bus is off - schedule is more than 10%.

Answer:

There is about a 3% chance that 5 or more buses are running off schedule. This is unusual and is convincing evidence that the true probability that a bus is off schedule is more than 10%.