Question

(a) An electrical firm manufactures light bulbs 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 bulbs 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 television sets in the city of Lusaka, Zambia,
it was found that 340 owned colour sets. Find a 95% confidence interval for the actual
proportion of families in this city with colour sets.

Answer 1

$a) Given \; that, μ=800, σ=40, n=16, then,\\ P(x<775)=P(Z<\frac{775-800}{\frac{40}{\sqrt{16}}})\\ =P(Z<-2.5)=0.5-P(0<Z<2.5)\\ =0.5-0.4938=0.0062\\ b) \text{The confidence interval for the mean }\\ \text{can be calculated as follows:}\\ \bar x±Z_{\frac{α}{2}}(\frac{s}{\sqrt{n}}),\\ \bar x=\frac{9.8+10.2+10.4+9.8+10.0+10.2+9.6}{7}\\ =\frac{70}{7}=10,\\ Z_{\frac{α}{2}}=Z_{\frac{0.05}{2}}=Z_{0.025}=1.96,\\ s^2=\frac{∑(x_{i}-x)^2}{n-1}\\ =\frac{(9.8-10)^2+(10.2-10)^2+...+(9.6-10)^2}{6}\\ =\frac{0.48}{6}=0.08,\\ s=\sqrt{s^2},\\ \text{ the lower limit }= 10-1.96(\frac{\sqrt{0.08}}{\sqrt{7}})\\ ≈9.79,\\ \text{ the upper limit} = 10+1.96(\frac{\sqrt{0.08}}{\sqrt{7}})\\ ≈10.21,\\ \text{so the confidence interval is}\\ 9.79\leq \bar x \leq 10.21\\ c) \text{The confidence interval for the actual }\\ \text{proportion can be calculated as follows:}\\ \hat p ±Z_{\frac{α}{2}}(\sqrt {\frac{p (1-p )}{n}}),\\ \hat p =\frac{340}{500}=0.68,\\ Z_{\frac{α}{2}}=Z_{\frac{0.05}{2}}=Z_{0.025}=1.96,\\ \text{the lower limit }= 0.68-1.96(\sqrt {\frac{0.68 (1-0.68 )}{500}})\\ ≈0.639,\\ \text{the upper limit }= 0.68+1.96(\sqrt {\frac{0.68 (1-0.68 )}{500}})\\ ≈0.721,\\ \text{so the confidence interval is}\\ 0.639\leq \hat p \leq 0.721$