4.40 CFLBs: A manufacturer of compact fluorescent light bulbs advertises that the distribution of the
lifespans of these light bulbs is nearly normal with a mean of 9,000 hours and a standard deviation of 1,000
hours.
(a) What is the probability that a randomly chosen light bulb lasts more than 10,500 hours?
(please round to four decimal places) (b) Describe the distribution of the mean
lifespan of 15 light bulbs.
right skewed
approximately normal with and
approximately normal with and
hours?
What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500
(please round to four decimal places)

Ask by Olson Wood. in the United States

Mar 27,2025

Question

4.40 CFLBs: A manufacturer of compact fluorescent light bulbs advertises that the distribution of the
lifespans of these light bulbs is nearly normal with a mean of 9,000 hours and a standard deviation of 1,000
hours.
(a) What is the probability that a randomly chosen light bulb lasts more than 10,500 hours?
(please round to four decimal places) (b) Describe the distribution of the mean
lifespan of 15 light bulbs.
right skewed
approximately normal with and
approximately normal with and
hours?
What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500
(please round to four decimal places)

Ask by Olson Wood. in the United States

Mar 27,2025

UpStudy AI · Accepted Answer

(a) Let $$ X $$ be the lifespan of a randomly chosen light bulb. We are given that 
$$
X \sim N(9000, 1000).
$$
To find the probability that a light bulb lasts more than 10,500 hours, calculate the z‐score:
$$
z = \frac{10500 - 9000}{1000} = 1.5.
$$
Thus, the probability is:
$$
P(X > 10500) = 1 - \Phi(1.5) \approx 1 - 0.9332 = 0.0668.
$$

(b) For the mean lifespan of $$ n = 15 $$ light bulbs, let $$\bar{X}$$ be the sample mean. The Central Limit Theorem tells us that when the individual lifespans are normally distributed, the sample mean is also normally distributed:
$$
\bar{X} \sim N\left(9000, \frac{1000}{\sqrt{15}}\right).
$$
So the correct description is: approximately normal with $$\mu = 9000$$ and $$\sigma = \frac{1000}{\sqrt{15}}$$.

To find the probability that the mean lifespan of 15 light bulbs is more than 10,500 hours, compute the z‐score for $$\bar{X}$$:
$$
z = \frac{10500 - 9000}{1000/\sqrt{15}} = \frac{1500 \sqrt{15}}{1000}.
$$
Approximating $$\sqrt{15} \approx 3.873$$, we have:
$$
z \approx \frac{1500 \times 3.873}{1000} \approx 5.809.
$$
The probability is:
$$
P(\bar{X} > 10500) = 1 - \Phi(5.809).
$$
Since a z-score of 5.809 is extremely high, this probability is essentially
$$
\approx 0.0000.
$$

Thus, the answers are:

(a) $$P(X > 10500) \approx 0.0668.$$

(b) The distribution of the mean lifespan is approximately normal with $$\mu = 9000$$ and $$\sigma = \frac{1000}{\sqrt{15}}$$, and $$P(\bar{X} > 10500) \approx 0.0000$$.
(a) The probability that a randomly chosen light bulb lasts more than 10,500 hours is approximately 0.0668. (b) The distribution of the mean lifespan of 15 light bulbs is approximately normal with a mean of 9,000 hours and a standard deviation of approximately 258.19 hours. The probability that the mean lifespan of 15 light bulbs is more than 10,500 hours is approximately 0.0000.

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