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avanturin [10]
2 years ago
6

Sofia's audio player has 15,000 songs. The play time for the songs is skewed to the right, with a mean of 255 seconds and a stan

dard deviation of 30 seconds. Part A: Can you accurately calculate the probability that the mean play time is more than 260 seconds for an SRS of 15 songs? Explain. (4 points) Part B: If you take a random sample of 40 songs instead of 15, explain how the Central Limit Theorem allows you to find the probability that the mean play time is more than 260 seconds. Calculate this probability and show your work. (6 points)
Mathematics
1 answer:
erma4kov [3.2K]2 years ago
7 0

Answer:

14.59%.

Step-by-step explanation:

Part A

You can not accurately calculate this because the SRS is not over 30. The CLT states that if the sample size is big enough, it will have a normal distribution. 15 is not big enough.

Part B

You can calculate this because the sample size is over 30; in this case, it is 40. So first, we have to find the standard deviation of the sampling distribution of the means. Which is done by taking the standard deviation and dividing it by the square root of the sample size, which comes out to be 4.74341649. Next, we throw it into the calculator in Normal CDF (260,9999,255,4.74341649). The final answer comes out to be 14.59%.

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Derek has 12 shirt in his closet. If 2 out of every 3 of these shirts are striped,how many striped shirts does Derek have in his
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12 divided by 3 is 4
if 2 of each of these is stripped that means 8/12 are striped
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Maria will spin the arrow on the spinner 2 times. What is the probability that the arrow will stop on the same letter twice? 19
gavmur [86]
Suppose the spinner lands on <em>a</em>.  There's a 1/3 chance that it'll land on <em>a</em> the second time.
Suppose the spinner lands on <em>b</em>.  There's a 1/3 chance that it'll land on <em>b</em> the second time.
Suppose the spinner lands on <em>c</em>.  There's a 1/3 chance that it'll land on <em>c</em> the second time.

We've covered all possibilities for the first spin, and they're all equal, so their average is 1/3.

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2 years ago
A recent poll of 124 randomly selected residents of a town with a population of 310 showed that 93 of them are opposed to a new
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The sample size is 124.
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So, 
n = 124
p = \frac{93}{124}=0.75

The point estimate of the population proportion = p = 0.75
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Margin of error (E) can be calculated by:

E= Z_{c}  \sqrt{ \frac{pq}{n} }

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Three pounds of squid can be purchased at the market for $18. Determine the equation and represent the function that defines the
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From the information given,

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A brewery produces cans of beer that are supposed to contain exactly 12 ounces. But owing to the inevitable variation in the fil
MArishka [77]

Answer:

T \sim N (\mu = 6*12=72 , \sigma= \sqrt{6} *0.3=0.735)

P(T \leq 72) = P(Z< \frac{72-72}{0.735}) = P(Z

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solutio to the problem

Let X the random variable that represent the amount of beer in each can of a population, and for this case we know the distribution for X is given by:

X \sim N(12,0.3)  

Where \mu=12 and \sigma=0.3

For this case we select 6 cans and we are interested in the probability that the total would be less or equal than 72 ounces. So we need to find a distribution for the total.

The definition of sample mean is given by:

\bar X = \frac{\sum_{i=1}^n X_i}{n} = \frac{T}{n}

If we solve for the total T we got:

T= n \bar X

For this case then the expected value and variance are given by:

E(T) = n E(\bar X) =n \mu

Var(T) = n^2 Var(\bar X)= n^2 \frac{\sigma^2}{n}= n \sigma^2

And the deviation is just:

Sd(T) = \sqrt{n} \sigma

So then the distribution for the total would be also normal and given by:

T \sim N (\mu = 6*12=72 , \sigma= \sqrt{6} *0.3=0.735)

And we want this probability:

P(T\leq 72)

And we can use the z score formula given by:

z = \frac{x-\mu}{\sigma}

P(T \leq 72) = P(Z< \frac{72-72}{0.735}) = P(Z

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