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VashaNatasha [74]
3 years ago
13

If a light bulb manufacturing company wants to​ estimate, with 99​% ​confidence, the mean life of compact fluorescent light bulb

s to within plus or minus175 hours and also assumes that the population standard deviation is 1000 ​hours, how many compact fluorescent light bulbs need to be​ selected?
Mathematics
2 answers:
matrenka [14]3 years ago
6 0

Answer:

217

Step-by-step explanation:

Reptile [31]3 years ago
4 0

Answer:

We need at least 217 compact fluorescent light bulbs

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.99}{2} = 0.005

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.005 = 0.995, so z = 2.575

Now, find M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

How many compact fluorescent light bulbs need to be​ selected?

We need at least n bulbs, in which n is found when M = 175, \sigma = 1000

So

M = z*\frac{\sigma}{\sqrt{n}}

175 = 2.575*\frac{1000}{\sqrt{n}}

175\sqrt{n} = 2575

\sqrt{n} = \frac{2575}{175}

\sqrt{n} = 14.71

\sqrt{n}^{2} = (14.71)^{2}

n = 216.5

We need at least 217 compact fluorescent light bulbs

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Answer:

S = 8

Step-by-step explanation:

An infinite geometric series is defined as limit of partial sum of geometric sequences. Therefore, to find the infinite sum, we have to find the partial sum first then input limit approaches infinity.

However, fortunately, the infinite geometric series has already set up for you. It’s got the formula for itself which is:

\displaystyle \large{\lim_{n\to \infty} S_n = \dfrac{a_1}{1-r}}

We can also write in summation notation rather S-term as:

\displaystyle \large{\sum_{n=1}^\infty a_1r^{n-1} = \dfrac{a_1}{1-r}}

Keep in mind that these only work for when |r| < 1 or else it will diverge.

Also, how fortunately, the given summation fits the formula pattern so we do not have to do anything but simply apply the formula in.

\displaystyle \large{\sum_{n=1}^\infty 4(0.5)^{n-1} = \dfrac{4}{1-0.5}}\\\\\displaystyle \large{\sum_{n=1}^\infty 4(0.5)^{n-1} = \dfrac{4}{0.5}}\\\\\displaystyle \large{\sum_{n=1}^\infty 4(0.5)^{n-1} = 8}

Therefore, the sum will converge to 8.

Please let me know if you have any questions!

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2 years ago
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Answer:

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Step-by-step explanation:

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