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:4.924 yearsStep-by-step explanation:Lets denote X the lifetime of a tv tube (In years). X has distribution [tex] N(\mu, 1.1) [/tex] , with [tex] \mu [/tex] unknown. We know that P(X < 4) = 0.2. Using this data, we can find the value of [tex] \mu [/tex] throught standarization.Lets call [tex] Z = \frac{X - \mu}{1.1} [/tex] the standarization of X. Z has distribution N(0,1), and its cummulative function, [tex] \phi [/tex] is tabulated. The values of [tex] \phi [/tex] can be found in the attached file. [tex] 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 [/tex]The value q such that [tex] \phi (q) = 0.2 [/tex] 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 [tex] \phi(-q) = 0.8 [/tex] , hence -q must be equal to 0.84. Thus, q = -0.84 But this value of q should match with the number [tex] \frac{4 - \mu}{1.1} [/tex] , so we have[tex] \frac{4- \mu}{1.1} = -0.84 [/tex][tex] 4 - \mu = 1.1 * (-0.84) = -0.924 [/tex][tex] \mu = 4 - (-0.924) = 4.924 [/tex]Thus, the expected lifetime of TV tubes is 4.924 years.I hope this works for you!