# Exercise to calculate the normal distribution of a light bulb companyNormal Distribution Example

### Problem Data

It is required to calculate the probability of P(900<X<1300) (interval) with the following data:

Mean (μ)
1200
Standard Deviation (σ)
250
Description
Exercise to calculate the normal distribution of a light bulb companyA company produces light bulbs whose lifetime follows a normal distribution that has a mean of 1,200 hours and a standard deviation of 250 hours. If we choose a light bulb randomly, what is the probability that it lasts between 900 and 1,300 hours?

## Solution

Below are the detailed calculations and graphs to obtain the probability according to the data provided:

### Step 1:

We have the normal random variable X with mean μ and variance σ2; that is, X ~ N(μ,σ2), which will become the standard normal random variable, Z ~ N(0,1). To do so, we will apply the following transformation:

We have μ = 1200 and σ = 250, then:

The exact result of the expression obtained is:

To check the result using the lower-tailed normal distribution (Z) table, we will make use of the following property:

In our problem it would be:

1. For P(Z<0.4)

The exact result of the expression obtained is:

To check the result we will use the left-tailed normal distribution (Z) table.

We will use our table to find the probability of P(Z<0.40) as follows:

• In the rows of the table we place the value of 0.4
• In the columns of the table we place the value of 0.00
• The intersection of the indicated row and column will give us the value of P(Z<0.40)=0.6554 (see highlighted value in the table below).

2. For P(Z<-1.2)

The exact result of the expression obtained is:

To check the result we will use the left-tailed normal distribution (Z) table.

Since our table does not show negative values, we will use the symmetry of the normal distribution to solve the problem:

We will use our table to find the probability of P(Z<1.20) as follows:

• In the rows of the table we place the value of 1.2
• In the columns of the table we place the value of 0.00
• The intersection of the indicated row and column will give us the value of P(Z<1.20)=0.8849 (see highlighted value in the table below).

Finally, we replace the value in the obtained expression:

Returning to the original expression, we replace the values obtained:

The probability that the random variable X lies in the interval [900,1300] is 0.5404

Note: The results obtained by the calculator and the one obtained by the table may have small differences due to rounding issues.

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