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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