Question

Suppose that the lifetime of tv tubes are normally distributed with a standard deviation of 1.1 years. Suppose that exactly 20% of tubes die before four years. Find the mean lifetime of TV tubes?

Answer 1

Answer:

4.924 years

Step-by-step explanation:

Lets denote X the lifetime of a tv tube (In years). X has distribution $N(\mu, 1.1)$ , with $\mu$ unknown.

We know that P(X < 4) = 0.2. Using this data, we can find the value of $\mu$ throught standarization.

Lets call $Z = \frac{X - \mu}{1.1}$ the standarization of X. Z has distribution N(0,1), and its cummulative function, $\phi$ is tabulated. The values of $\phi$ can be found in the attached file.

$P(X < 4) = P(\frac{X - \mu}{1.1} < \frac{4 - \mu}{1.1}) = P(Z < \frac{4 - \mu}{1.1}) = \phi(\frac{4 - \mu}{1.1}) = 0.2$

The value q such that $\phi (q) = 0.2$ doesnt appear on the table. We can find it by using the symmetry of the normal density function. The opposite of q, -q must verify that $\phi(-q) = 0.8$ , hence -q must be equal to 0.84. Thus, q = -0.84

But this value of q should match with the number $\frac{4 - \mu}{1.1}$ , so we have

$\frac{4- \mu}{1.1} = -0.84$

$4 - \mu = 1.1 * (-0.84) = -0.924$

$\mu = 4 - (-0.924) = 4.924$

Thus, the expected lifetime of TV tubes is 4.924 years.

I hope this works for you!

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