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coldgirl [10]
3 years ago
6

Judy just calculated the standard deviation for a project she is working on. The value for the standard deviation is a small num

ber. What does this represent?
The data is spread far from the mean.

The data is spread close to the mean.

The data is spread equal to the mean.

None of these choices are correct.
Mathematics
1 answer:
Contact [7]3 years ago
7 0

Answer: b) The data is spread close to the mean

Step-by-step explanation: Theoretically, when the standard deviation of a statistical data is large or high, it simply means that the values in such statistical data set are farther away from the mean, On the other hand when the standard deviation is small or low, it simply means that the values of such statistical data are close to the mean of those data on average.

Therefore Judy's low standard deviation for her project simply indicates that her statistical data set is spread close to the mean.

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Answer:

800 flyersOn Earth day, a local community wants to distribute 800 flyers, 300 banners, and 250 badges to volunteers in packets containing the three items. What is the greatest number of packets that can be made using all these items if each type of item is equally distributed among the packets?

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3 years ago
The manager wants to advertise that anybody who isn't served within a certain number of minutes gets a free hamburger. But she d
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Answer:

The number of minutes advertisement should use is found.

x ≅ 12 mins

Step-by-step explanation:

(MISSING PART OF THE QUESTION: AVERAGE WAITING TIME = 2.5 MINUTES)

<h3 /><h3>Step 1</h3>

For such problems, we can use probability density function, in which probability is found out by taking integral of a function across an interval.

Probability Density Function is given by:

f(t)=\left \{ {{0 ,\-t

Consider the second function:

f(t)=\frac{e^{-t/\mu}}{\mu}\\

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The function f(t) becomes

f(t)=0.4e^{-0.4t}

<h3>Step 2</h3>

The manager wants to give free hamburgers to only 1% of her costumers, which means that probability of a costumer getting a free hamburger is 0.01

The probability that a costumer has to wait for more than x minutes is:

\int\limits^\infty_x {f(t)} \, dt= \int\limits^\infty_x {}0.4e^{-0.4t}dt

which is equal to 0.01

<h3>Step 3</h3>

Solve the equation for x

\int\limits^{\infty}_x {0.4e^{-0.4t}} \, dt =0.01\\\\\frac{0.4e^{-0.4t}}{-0.4}=0.01\\\\-e^{-0.4t} |^\infty_x =0.01\\\\e^{-0.4x}=0.01

Take natural log on both sides

ln (e^{-0.4x})=ln(0.01)\\-0.4x=ln(0.01)\\-0.4x=-4.61\\x= 11.53

<h3>Results</h3>

The costumer has to wait x = 11.53 mins ≅ 12 mins to get a free hamburger

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