Answer to Question #120496 in Statistics and Probability for FRED

Question

(a) An electrical firm manufactures fire balls that have a length of life that is approximately
normal distributed with mean equal to 800 hours and a standard deviation of 40 hours. Find
the probability that a random sample of 16 balls will have an average life of less than 775
hours.
(b) The contents of seven similar containers of medicine are 9.8, 10.2, 10.4, 9.8, 10.0, 10.2 and
9.6 litres. Find a 95% confidence interval for the mean of all such containers, assuming an
approximate normal distribution
(c) In a random sample of 500 families owning an extingiusher sets in the city of Lusaka, Zambia,
it was found that 340 owned dry powder sets. Find a 95% confidence interval for the actual
proportion of families in this city with dry powder fire extinguisher sets.

Answer 1

a) Let $X=$ the length of life: $X\sim N(\mu,\sigma^2/n).$ Then $Z=\dfrac{X-\mu}{\sigma/\sqrt{n}}\sim N(0,1)$

Given $\mu=800\ h, \sigma=40\ h,n=16.$

$$P(X<775)=P(Z<{775-800\over 40/\sqrt{16}})=P(Z<-2.5)\approx$$ $$\approx0.0062$$

b)

$$\bar{x}={1\over n}\displaystyle\sum_{i=1}^nx_i=10.0$$

$$s^2={1\over n-1}\displaystyle\sum_{i=1}^n(x_i-\bar{x})^2=0.08$$

$$s=\sqrt{s^2}\approx0.282843$$

Based on the provided information, the critical t-value for $\alpha=0.05$ and $df=7-1=6$degrees of freedom is $t_c=2.447.$  

The 95% confidence for the population mean $\mu$ is computed using the following expression

$$CI=(\bar{x}-t_c\cdot{s\over \sqrt{n}}, \bar{x}+t_c\cdot{s\over \sqrt{n}})=$$

$$=(10.0-2.447\cdot{0.282843\over \sqrt{7}}, 10.0+2.447\cdot{0.282843\over \sqrt{7}})\approx$$

$$\approx(9.7384,10.2616)$$

The 95 % confidence for the population mean $\mu$  is $(9.7384,10.2616)$

$$9.7384≤μ≤10.2616$$

c) $\hat{p}=\dfrac{X}{N}=\dfrac{340}{500}=0.68$

The critical value for $\alpha=0.05$ is $z_c=z_{1-\alpha/2}=1.96.$

The corresponding confidence interval is computed as shown below:

$$CI=( \hat{p}-z_c\sqrt{{\hat{p}(1-\hat{p})\over n}} , \hat{p}+z_c\sqrt{{\hat{p}(1-\hat{p})\over n}} )=$$

$$=(0.68-1.96\sqrt{{0.68(1-0.68)\over 500}} , 0.68+1.96\sqrt{{0.68(1-0.68)\over 500}} )\approx$$

$$\approx(0.6391, 0.7209)$$

Therefore, based on the data provided, the 95% confidence interval for the population proportion is $0.6391<p< 0.7209,$ which indicates that we are 95% confident that the true population proportion $p$ is contained by the interval $(0.6391, 0.7209).$