# 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 company**A 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 σ^{2}; that is, ** X ~ N(μ,σ^{2})**, which will become the standard normal random variable,

**. To do so, we will apply the following transformation:**

*Z ~ N(0,1)*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
(see highlighted value in the table below).*P(Z<0.40)=0.6554*

**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
(see highlighted value in the table below).*P(Z<1.20)=0.8849*

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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