Question-24
The average life of a coffee machine is 5 years, with a standard deviation of 1 year. Assuming that the lives of these machines follow approximately a normal distribution, find the probability that the mean life of a random sample of 16 such machines falls between 4.6 and 5.4 years. Enter your answer correct to two decimals.
If \(X \sim N(\mu, \sigma^2)\), then \(\overline{X} \sim N\left(\mu, \dfrac{\sigma^2}{n}\right)\), where \(\overline{X} = \dfrac{\sum_{i=1}^{n}x_i}{n}\).
Also, \(Z = \dfrac{X - E(X)}{\sqrt{Var(X)}} \sim N(0,1)\) is a standard normal variate.
Use \(F_Z(1.6) = 0.9452\)
0.89
According to the question, we have \(\mu = 5\) and \(\sigma = 1\).
Let \(X_i\) represents the life of the \(i^{th}\) machine, then \(X_i \sim N\left(5, 1\right)\), \(\forall \hspace{1mm}i=1, 2, \ldots, 16\).
Now,
\(\overline{X} \sim N\left(5, \dfrac{1}{16}\right)\), where \(\overline{X} = \dfrac{\sum_{i=1}^{16}x_i}{16}\) will represent the mean life of a random sample of 16 machines.
\(P(4.6 < \overline{X} < 5.4) = P\left(\dfrac{4.6-5}{1/4} < \dfrac{\overline{X}-5}{1/4} < \dfrac{5.4-5}{1/4}\right)\)
\(=P\left(\dfrac{-0.4}{1/4} < Z < \dfrac{0.4}{1/4}\right)\)
\(= P\left(-1.6 < Z < 1.6\right)\)
\(= P\left(Z < 1.6\right) - P\left(Z < -1.6\right)\)
Since, \(Z\) is a standard normal variable, therefore \(P(Z<-a) = P(Z>a)\). Thus, \(P(Z<-1.6) = P(Z>1.6)\). Now,
\(= P\left(Z < 1.6\right) - P\left(Z > 1.6\right)\)
\(= P\left(Z < 1.6\right) - \left(1-P\left(Z < 1.6\right)\right)\)
\(= P\left(Z < 1.6\right) - 1 + P\left(Z < 1.6\right)\)
\(= 2P\left(Z < 1.6\right) - 1\)
\(= 2F_Z(1.6) - 1\)
\(= 2(0.9452) - 1 = 0.89 \hspace{2mm}\text{(till two decimal places.)}\)