Question

The physical plant at the main campus of a large state university receives daily requests to replace fluorescent lightbulbs. The distribution of the number of daily requests is approximately normal and has a mean of 62 and a standard deviation of 5. Use the Empirical Rule to determine the approximate proportion of lightbulb replacement requests numbering between 62 and 72?

Answer 1

`P(62< X< 72)= P(X<72) -P(X<62)=0.975-0.5=0.475`

Explanation:

The empirical rule, also known as three-sigma rule or 68-95-99.7 rule, "is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)".

Let X the random variable who represent the courtship time (minutes).

From the problem we have the mean and the standard deviation for the random variable X.
`E(X)=62, Sd(X)=5`

So we can assume
`\mu=62 , \sigma=5`

On this case in order to check if the random variable X follows a normal distribution we can use the empirical rule that states the following:

• The probability of obtain values within one deviation from the mean is 0.68

• The probability of obtain values within two deviation's from the mean is 0.95

• The probability of obtain values within three deviation's from the mean is 0.997

So we need values such that

`P(X<\mu -\sigma)=P(X <57)=0.16`

`P(X>\mu +\sigma)=P(X >67)=0.16`

`P(X<\mu -2*\sigma)=P(X<52)=0.025`

`P(X>\mu +2*\sigma)=P(X>72)=0.025`

`P(X<\mu -3*\sigma)=P(X<47)=0.0015`

`P(X>\mu +3*\sigma)=P(X>77)=0.0015`

For this case we want to find this probability:

`P(62 < X< 72)`

And we can find this probability on this way:

`P(62< X< 72)= P(X<72) -P(X<62)`

Since
`P(X>72) =0.025` by the complement rule we have that:

`P(X<72) = 1-0.025 =0.975`

And
`P(X<62) =0.5` because for this case 62 is the mean.

So then we have this:

`P(62< X< 72)= P(X<72) -P(X<62)=0.975-0.5=0.475`