Mathematics, 05.11.2020 19:10 SamanthaBees

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?

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Answer from: Tobymorgan3287

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

a) 0.5111 = 51.11% probability that a bulb burns between 778 and 834 hours.b) 0.0008 = 0.08% probability that the average light bulb life is at least 820 hours.Normal Probability Distribution

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

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean. After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X. By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

In this problem:

Mean of 800 hours, hence $\mu = 800$ .Standard deviation of 40 hours, hence $\sigma = 40$ .

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:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{834 - 800}{40}$

Z = 0.85

Z = 0.85 has a p-value of 0.8023.

X = 778:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{778 - 800}{40}$

Z = -0.55

Z = -0.55 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 $n = 40, s = \frac{40}{\sqrt{40}}$ .

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

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{820 - 800}{\frac{40}{\sqrt{40}}}$

Z = 3.16

Z = 3.16 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 link

Answer from: crybaby222psyc

(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:

P(778

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:

$P(\bar X\geq 820)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}\geq \frac{820-800}{40/\sqrt{40}})\\\\=P(Z3.16)\\\\=1-P(Z$

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

Answer from: Quest

6hope this ! ! ’

Answer from: Quest

None liner bro because
Will give is this a strong positive linear association or something else?

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