SCSA Mathematics Methods Discrete random variables

The sample proportion of heads is given by

$\hat{p} = \frac{30}{50} = 0.6$

Hence, the 90% confidence interval is

$0.6 - 1.645\sqrt{\frac{0.6 \times 0.4}{50}} \le p \le 0.6 + 1.645\sqrt{\frac{0.6 \times 0.4}{50}}$

$0.4860 \le p \le 0.7140$

$X \sim \text{Bin}(20, 0.9)$

The expected value of $X$ is given by

$E(X) = 20 \times 0.9$

$= 18$

The variance of $X$ is given by

$Var(X) = 20 \times 0.9 \times 0.1$

$= 1.8$

If three confidence intervals did not contain the true proportion, then 17 did contain the true proportion.

$\text{P}(X = 17) = 0.1901$

The expected value of a uniformly distributed continuous random variable $X$ is midway between the maximum and minimum values, so the probability density function is

$f(x)=\begin{cases} \dfrac{1}{6}, & 3\le x\le 9 \\ 0, & \text{otherwise} \end{cases}$

The variance of $X$ is given by

$\begin{aligned} \operatorname{Var}(X)&=\int_{3}^{9} (x-6)^2 f(x)\,dx\\ &=\dfrac{1}{6}\int_{3}^{9} (x-6)^2\,dx\\ &=\dfrac{1}{6}\left[\dfrac{(x-6)^3}{3}\right]_{3}^{9}\\ &=\dfrac{1}{6}\left(9-(-9)\right)\\ &=3 \end{aligned}$

From the question $np=\dfrac{1}{2}$ and $np(1-p)=\dfrac{5}{12}$. It follows that

$\begin{aligned} \dfrac{1}{2}(1-p)&=\dfrac{5}{12}\\ \Rightarrow 1-p&=\dfrac{5}{6}\\ \Rightarrow p&=\dfrac{1}{6} \end{aligned}$

And

$\begin{aligned} \dfrac{n}{6}&=\dfrac{1}{2}\\ \Rightarrow n&=3 \end{aligned}$

Hence,

$\begin{aligned} P(W=1)&=\binom{3}{1}\left(\dfrac{1}{6}\right)^1\left(\dfrac{5}{6}\right)^2\\ &=3\times\dfrac{1}{6}\times\dfrac{25}{36}\\ &=\dfrac{25}{72} \end{aligned}$

Let the random variable $X$ denote the number of new businesses that fail out of the 500. Then $X \sim Bin(500, 0.2)$. $P(X \le 120) = 0.9877$

Sample proportion $\hat{p} = N\left(0.2, \frac{0.2 \times 0.8}{500}\right)$,
That is, $\hat{p} = N(0.2, 0.00032)$, as the sample size is large

$P(\hat{p} < 0.18) = 0.1318$

Sample proportion $\hat{p} = \frac{90}{500} = 0.18$
95% confidence interval $\left( 0.18 - 1.96 \times \sqrt{\frac{0.18 \times 0.82}{500}}, \ 0.18 + 1.96 \times \sqrt{\frac{0.18 \times 0.82}{500}} \right)$
$= (0.1463, 0.2136)$

Sample proportion $\hat{p} = \frac{80}{500} = 0.16$
$E = 1.96 \times \sqrt{\frac{0.16 \times 0.84}{500}} = 0.0321$

Sample proportion $\hat{p} = \frac{80}{500} = 0.16$
95% confidence interval $\left( 0.16 - 1.96 \times \sqrt{\frac{0.16 \times 0.84}{500}}, \ 0.16 + 1.96 \times \sqrt{\frac{0.16 \times 0.84}{500}} \right)$
$= (0.1279, 0.1921)$
Comparing this confidence interval with the previous one $(0.1463, 0.2136)$, we can see that they overlap.
Therefore, it does not appear that the financial advice has reduced the proportion of businesses that fail in the first year.

The width of the confidence interval would be increased, as the margin of error of the sample proportion will increase, thus increasing the error.

1. We assume that the businesses fail independently of each other. This is unlikely to be valid because:
   (a) two similar businesses may both fail or both survive
   (b) if two similar businesses in an area then one may dominate and the other fails.
2. We assume that the probability of a business failing is the same for all businesses. This is unlikely to be valid, as businesses of different types are expected to have different probabilities of failing.
3. The probability of failure is fixed (for the year). This is unlikely to be true, and the probability will depend on changing conditions over time.