Question

3. A computing system manager states that the rate of interruptions to the internet

service is 0.2 per week. Use the Poisson distribution to find the probability of

a) one interruption in 3 weeks

b) at least two interruptions in 5 weeks

c) at most one interruption in 15 weeks

4. A special cable has an average breaking strength of 800 pounds. The standard deviation of 

the population is 12 pounds. A researcher selects a sample of 35 cables and founds that the 

average breaking strength is 805 pounds. Assume that the variable is normally distributed. 

A. Find the point estimate of the cable average breaking strength (μ) 

B. Construct the 95% confidence interval for the population average breaking strength of 

the cable (μ) and interpret your result. 

C. Can the researcher conclude that the average breaking strength of the cable greater than 

800 pounds at 1% level of significance? 

Answer 1

3

Let the random variable $X$ represent the number of interruptions of internet service. The rate of interruption to the internet service is $r=0.2\space per\space week$.

We determine these probabilities using Poisson distribution with parameter $\lambda$ given by,

$p(X=x)=e^{-\lambda}(\lambda)^x/x!$

a.

With $X=1,$ the rate $\lambda=r*3=0.2*3=0.6$

Therefore, $p(X=1)=e^{-0.6}*(\lambda)^1/1!=0.3293$

Thus, probability that there is one interruption in 3 weeks is 0.3293.

b.

With $X\geqslant2,$ parameter $\lambda=r*5=0.2*5=1$,

$p(X\geqslant2)=\displaystyle\sum p(X=x), \forall\space x\geqslant2$ which can be written as,

$1-\{p(X=0)+p(X=1)\}=1-\{0.3679+0.3679\}=1-0.7358=0.2642$

Hence probability that there are at least two interruptions in 5 weeks is 0.2642.

c.

With $X\leqslant$ 1, parameter $\lambda=r*15=0.2*15=3$

$p(X\leqslant1)=p(X=0)+p(X=1)=0.0498+0.1494=0.1991$

Probability that there is at most one interruption on 15 weeks is 0.1991.

4

a.

The point estimate for the population mean $(\mu)$ is the sample mean $(\bar{x})$. Hence, point estimate for $\mu$ is $\bar{x}=805.$

b.

$95\%$ confidence interval for the population average breaking strength of the cable$(\mu)$ is given as,

$C.I=\bar{x}\pm Z_{\alpha/2}*\sigma/\sqrt n$

From the data above,

$\bar{x}=805$

$Z_{\alpha/2}=Z_{0.05/2}=Z_{0.025}=1.96$

$\sigma=12$

$n=35$

With these values we can compute the $95\%$ as follows,

$C.I=805\pm1.96*(12/ \sqrt{35})$

$C.I=805\pm1.96*2.02837$

$C.I=805\pm3.9756$

$C.I=(801.02, 808.98)$

Since the population mean $\mu=800$ does not lie within this confidence interval, we reject the null hypothesis that the population mean is equal to 800 pounds.

c.

In this part, we test the hypothesis,

$H_0:\mu=800$

$Against$

$H_1:\mu\gt 800$

We first compute the test statistic given by,

$Z^*_c=(\bar{x}-\mu)/(\sigma/\sqrt{n})$

$Z^*_c=(805-800)/(12/\sqrt{35})$

$Z^*_c=5/2.02837=2.4650$ and compare with the table value at $\alpha=1\%$ level of significance.

The table value is $Z_{0.01}=2.33$ and we reject the null hypothesis if $Z^*_c\gt Z_{0.01}$

Since $Z^*_{c}=2.4650$ is greater than the table value, $Z_{0.01}=2.33,$ we reject the null hypothesis and conclude that sufficient evidence exist to show that mean breaking strength is greater than 800 pounds.