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Doss [256]
3 years ago
5

he physical plant at the main campus of a large state university receives daily requests to replace florescent light-bulbs. The

distribution of the number of daily requests is Normally distributed with a mean of 47 and a standard deviation of 10. Using the Empirical Rule, what is the approximate percentage of light-bulb replacement requests numbering between 47 and 57?
Mathematics
1 answer:
Stolb23 [73]3 years ago
8 0

Answer:

47.75 %

Step-by-step explanation:

It is a very well known issue that in Standard Normal Distribution  porcentages of all values fall according to:

μ  +   σ       will contain a 68.3 %

μ   +  2σ    will contain  a  95.5 %

μ  +  3σ      will contain  a  99.7 %

However it is extremely  importan to understand that the quantities above mentioned are distributed simmetrically at both sides of the mean, that is,  the intervals are:

[  μ - 0,5σ ;  μ  + 0,5σ ]

[  μ -   1σ ;  μ  +      1σ ]

[  μ -   1.5σ ;  μ  +   1.5σ ]

So we have to take that fact into account when applying the empirical rule. Then

With   mean    μ  =  47       and  σ = 10   is equal to say

values between    47    and     57    (  μ  +   σ  )  we are talking about the second interval, but just half of it.

Then the  approximate porcentage of light-bulb replacement requests is

95.5 /2   =  47.75 %

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The probability that the card is not yellow is \frac{5}{9}

<em><u>Solution:</u></em>

<em><u>The probability of an event is given as:</u></em>

Probability = \frac{ \text{ number of favorable outcomes }}{ \text{ total number of possible outcomes }}

From given,

A box is filled with 8 blue cards, 8 yellow cards, and 2 green cards

Total number of possible outcomes = 8 + 8 + 2 = 18

Favorable outcome = card is not yellow

Favorable outcome = card is blue or green = 8 + 2 = 10

Therefore,

Probability = \frac{10}{18}\\\\Probability = \frac{5}{9}

Thus the probability that the card is not yellow is 5/9

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MrRa [10]

Answer:

Step-by-step explanation:

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inserting the parameters

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Step-by-step explanation:

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A manufacturing company regularly conducts quality control checks at specified periods on the products it manufactures. Historic
Rina8888 [55]

Answer:

The probability that none of the LED light bulbs are​ defective is 0.7374.

Step-by-step explanation:

The complete question is:

What is the probability that none of the LED light bulbs are​ defective?

Solution:

Let the random variable <em>X</em> represent the number of defective LED light bulbs.

The probability of a LED light bulb being defective is, P (X) = <em>p</em> = 0.03.

A random sample of <em>n</em> = 10 LED light bulbs is selected.

The event of a specific LED light bulb being defective is independent of the other bulbs.

The random variable <em>X</em> thus follows a Binomial distribution with parameters <em>n</em> = 10 and <em>p</em> = 0.03.

The probability mass function of <em>X</em> is:

P(X=x)={10\choose x}(0.03)^{x}(1-0.03)^{10-x};\ x=0,1,2,3...

Compute the probability that none of the LED light bulbs are​ defective as follows:

P(X=0)={10\choose 0}(0.03)^{0}(1-0.03)^{10-0}

                =1\times 1\times 0.737424\\=0.737424\\\approx 0.7374

Thus, the probability that none of the LED light bulbs are​ defective is 0.7374.

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