Answer :
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:
- 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 [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]
- 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 [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
In this problem:
- Mean of 800 hours, hence [tex]\mu = 800[/tex].
- Standard deviation of 40 hours, hence [tex]\sigma = 40[/tex].
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