The local authorities in a certain city install 10,000 electric lamps in the streets of the city. If these lamps have an average life of 1000 burning hours with a standard deviation of 200 hours, assuming normal distribution, what number of lights might expected

Step-by-step explanation:

It is given:

[tex]\mu=1000[/tex], [tex]\sigma=200[/tex]

Let x be the random variable that denotes the number of hours the lamp lights up.

what number of lights might expected to fail in the first 800 burning hours

Answer: We are required to find the expected number of lights to fail in the first 800 burning hours.

First, we have to find the probability as:

P(Z < 800)

Using the z-score formula, we have:

Using the z-score formula, we have:

[tex]P(x <800)=P \left(z<\frac{800-1000}{200} \right )[/tex]

=P(z<-1)

Now using the standard normal table, we have:

P(x<800)=P(z<-1)=0.1587

Therefore, the expected number of lights to fail in the first 800 burning hours is:

[tex]\boldsymbol{0.1587 \times 10000=1587}[/tex]

After how many burning hours 10% of the lamps would be still burning?

We first need to find the z-value corresponding to area = 1- 0.10 = 0.90. Using the standard normal table, we have:

z(0.90)=1.28

Now using the z-score formula, we have:

[tex]z= \frac{x-\mu}{\sigma}[/tex]

[tex]1.28= \frac{x-1000}{200}[/tex]

[tex]1.28 \times 200= x-1000[/tex]

[tex]256= x-1000[/tex]

x= 1000+256

[tex]\boldsymbol{x= 1256 }[/tex]

Using the Zscore principle, the number of bulbs expected to Fail in the first 800 burning hours is 1587

Given the Parameters :

The number of Lamps that might be expected to fail in the first 800 burning hours :

Recall:

Zscore = (800 - 1000) / 200 = -1

Using the normal distribution table :

P(Z < -1) = 0.15866

The number of bulbs expected to fail :

0.15866 × 10000 = 1586.6

Hence, 1587 bulbs are expected to fail in the first 800 burning hours.

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