Question

A TV manufacturer knows that their TV's have a normally distributed lifespan, with a mean of 5.6 years, and standard deviation of 1 years.

If a TV is picked at random, 3% of the time its life will be less than how many years?


Give your answer to one decimal place.

Answer 1

3% of the time its life will be less than 3.7 years.

Explanation:

We are given that a TV manufacturer knows that their TV's have a normally distributed lifespan, with a mean of 5.6 years, and standard deviation of 1 years.

Let X = Lifespan of TV's

So, X ~ Normal(
`\mu=5.6,\sigma^(2)=1^(2)`)

The z score probability distribution for normal distribution is given by;

Z =
`\frac{ X-\mu}{{\sigma} }` ~ N(0,1)

where,
`\mu` = population mean lifespan = 5.6 years

`\sigma` = standard deviation = 1 year

Now, if a TV is picked at random, we have to find that 3% of the time its life will be less than how many years, that means;

P(X < x) = 0.03 {where x is the required years}

P(
`\frac{ X-\mu}{{\sigma} }` <
`\frac{ x-5.6}{{1} }` ) = 0.03

P(Z <
`\frac{ x-5.6}{{1} }` ) = 0.03

Now, in the z table the critical value of x which represents the below 3% probability area is given as -1.881, i.e;

`\frac{ x-5.6}{{1} }` = -1.881

x = 5.6 - 1.881

x = 3.7

Hence, 3% of the time its life will be less than 3.7 years.