Answered

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18. Light Bulbs: The mean lifespan of a standard 60 watt incandescent light bulb is 875 hours with a standard deviation of 80 hours. The mean lifespan of a standard 14 watt compact fluorescent light bulb (CFL) is 10,000 hours with a standard deviation of 1,500 hours. These two bulbs put out about the same amount of light. Assume the lifespan’s of both types of bulbs are normally distributed to answer the following questions. (a) I select one incandescent light bulb and put it in my barn. It seems to last forever and I estimate that it has lasted more than 2000 hours. What is the probability of selecting a random incandescent light bulb and having it last 2000 hours or more. Did something unusual happen here? (b) I select one CFL bulb and put it in the bathroom. It doesn’t seem to last very long and I estimate that it has lasted less than 5,000 hours. What is the probability of selecting a random CFL and having it last less than 5,000 hours. Did something unusual happen here? (c) Compare the the lifespan of the middle 99% of all incandescent and CFL light bulbs. (d) Is there much of a chance that I happen to buy an incandescent light bulb that lasts longer than a randomly selected CFL?

Sagot :

Set

[tex]\begin{gathered} \mu_1=875,\sigma_1=80 \\ \text{and} \\ \mu_2=10000,\sigma=1500 \end{gathered}[/tex]

a) The Z-score formula is

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

Therefore, in our case, if x=2000

[tex]\Rightarrow Z=\frac{2000-875}{80}=\frac{1125}{80}=14.0625[/tex]

Using a Z-score table,

[tex]\begin{gathered} P(z\ge2000)=1-P(z<2000)\approx1-1=0 \\ \Rightarrow P(z\ge2000)=0 \end{gathered}[/tex]

A value of 2000 hrs is 14 standard deviations away from the mean. The probability is practically zero.

b) Similarly, set x=5000; then,

[tex]Z=\frac{5000-10000}{1500}=-\frac{5000}{1500}=-3.333\ldots[/tex]

Thus, using a z-score table

[tex]P(z\le5000)=0.0004[/tex]

The probability is 0.0004=0.04%. It is quite improbable but not an impossible event.

c) According to the empirical rule 99.8% of the data lies within 3 standard deviations; thus,

[tex]\begin{gathered} Incandescent \\ \mu_1\pm3\sigma_1=\lbrack875-240,875+240\rbrack=\lbrack635,1115\rbrack \\ \text{CFL} \\ \mu_2\pm3\sigma_2=\lbrack10000-4500,10000+4500\rbrack=\lbrack5500,14500\rbrack \end{gathered}[/tex]

The lifespan of 99% of all incandescent bulbs is between 635 and 1115 hrs, whereas that of all CFL bulbs is between 5500 and 14500 hrs.

d) If we randomly select a CFL, the most probably lifespan is the mean of the distribution, in other words, 10000 hrs.

The probability of an incandescent bulb lasting 10000 hrs is

[tex]\begin{gathered} Z=\frac{10000-875}{80}=114.0625 \\ \Rightarrow P(z\ge10000)=1-P(z<10000)=0 \end{gathered}[/tex]

The event is practically impossible.

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