An electrical firm manufactures light bulbs that have a life, before burn-out, that is normally distributed with mean of 800 hours and standard deviation of 40 hours.
A. Find the probability that a bulb burns between 778 and 834 hours.
B. A random sample of 40 bulbs from that firm is selected. What is the probability that the average light bulb life is at least 820 hours?

Answer:

(A) The probability that a bulb burns between 778 and 834 hours is 0.51118.

(B) The probability that the average light bulb life is at least 820 hours in a random sample of 40 bulbs  is 0.00079

Step-by-step explanation:

Let X denote the life time of light bulbs.

It is provided that X follows a normal distribution with mean of 800 hours and standard deviation of 40 hours.

(A)

Compute the probability that a bulb burns between 778 and 834 hours as follows:

[tex]P(778<X<834)=P(\frac{778-800}{40}<\frac{X-\mu}{\sigma}<\frac{834-800}{40})\\\\=P(-0.55<Z<0.85)\\\\=P(Z<0.85)-P(Z<-0.55)\\\\=0.80234-0.29116\\\\=0.51118[/tex]

Thus, the probability that a bulb burns between 778 and 834 hours is 0.51118.

(B)

Compute the probability that the average light bulb life is at least 820 hours in a random sample of 40 bulbs as follows:

[tex]P(\bar X\geq 820)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}\geq \frac{820-800}{40/\sqrt{40}})\\\\=P(Z>3.16)\\\\=1-P(Z<3.16)\\\\=1-0.99921\\\\=0.00079[/tex]

Thus, the probability that the average light bulb life is at least 820 hours in a random sample of 40 bulbs  is 0.00079

Using the normal distribution and the central limit theorem, it is found that there is a:

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

In this problem:

Item a:

The probability is the p-value of Z when X = 834 subtracted by the p-value of Z when X = 778, hence:

X = 834:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{834 - 800}{40}[/tex]

[tex]Z = 0.85[/tex]

[tex]Z = 0.85[/tex] has a p-value of 0.8023.

X = 778:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{778 - 800}{40}[/tex]

[tex]Z = -0.55[/tex]

[tex]Z = -0.55[/tex] has a p-value of 0.2912.

0.8023 - 0.2912 = 0.5111

0.5111 = 51.11% probability that a bulb burns between 778 and 834 hours.

Item b:

Sample of 40, hence [tex]n = 40, s = \frac{40}{\sqrt{40}}[/tex].

The probability is 1 subtracted by the p-value of Z when X = 820, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{820 - 800}{\frac{40}{\sqrt{40}}}[/tex]

[tex]Z = 3.16[/tex]

[tex]Z = 3.16[/tex] has a p-value of 0.9992.

1 - 0.9992 = 0.0008

0.0008 = 0.08% probability that the average light bulb life is at least 820 hours.

To learn more about the normal distribution and the central limit theorem, you can take a look at https://brainly.com/question/24663213

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