Question-25
An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed, with mean equal to 600 hours and a standard deviation of 50 hours. Find the probability that a random sample of 25 bulbs will have an average life of less than 615 hours. 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 \(F_Z(1.5) = 0.933\)
0.93
According to the question, we have \(\mu = 600\) and \(\sigma = 50\).
Let \(X_i\) represents the life of the \(i^{th}\) bulb, then \(X_i \sim N\left(600, 50\right)\), \(\forall \hspace{1mm}i=1, 2, \ldots, 25\).
Now,
\(\overline{X} \sim N\left(600, \dfrac{50^2}{25}\right)\), where \(\overline{X} = \dfrac{\sum_{i=1}^{25}x_i}{25}\) will represent the mean life of a random sample of 25 light bulbs.
\(P( \overline{X} < 615) = P\left(\dfrac{\overline{X}-600}{50/5} < \dfrac{615-600}{50/5}\right)\)
\(=P\left( Z < \dfrac{15}{10}\right)\)
\(= P\left(Z < 1.5\right)\)
\(= 0.93 \hspace{2mm}\text{(till two decimal places.)}\)