Answer to Question #286996 in Statistics and Probability for Aman

Question

In a sample of 1000 tube lights, the mean life and standard deviation of lives are 4500 hrs. And 1500 hrs. Respectively. Assuming the distribution to be normal, found the number of tubes lasting between 4000 hrs and 6000 hrs.

Answer 1

Let $X=$ the variable representing the life of tube lights: $X\sim N(\mu, \sigma^2).$

Given $\mu=4500 h, \sigma=1500h.$

$$P(4000<X<6000)=P(X<6000)-P(X\leq 4000)$$

$$=P(Z<\dfrac{6000-4500}{1500})-P(Z\leq \dfrac{4000-4500}{1500})$$

$$=P(Z<1)-P(Z\leq -\dfrac{1}{3})$$

$$\approx0.8413447-0.3694413$$

$$\approx0.4719$$

$$1000(0.4719)=472$$

The number of tubes lasting between 4000 hrs and 6000 hrs is $472.$