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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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Answer:

9.9676 - 2.326*0.5904 =8.594

9.9676 + 2.326*0.5904 =11.341

Step-by-step explanation:

Notation

\bar X represent the sample mean

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

For this case the 9% confidence interval is given by:

8.8104 \leq \mu \leq 11.1248

We can calculate the mean with the following:

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The margin of error for this case is given by:

ME = t_{\alpha/2}\frac{s}{\sqrt{n}} = t_{\alpha/2} SE

And we can solve for the standard error:

SE = \frac{ME}{t_{\alpha/2}}

The critical value for 95% confidence using the normal standard distribution is approximately 1.96 and replacing we got:

SE = \frac{1.1572}{1.96}= 0.5904

Now for the 98% confidence interval the significance is \alpha=1-0.98= 0.02 and \alpha/2 = 0.01 the critical value would be 2.326 and then the confidence interval would be:

9.9676 - 2.326*0.5904 =8.594

9.9676 + 2.326*0.5904 =11.341

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the statements given above are true.

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He spun a 4 five times.

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iv) If the number of trials is changed, the predicted number of outcomes also changes.

v) If the number of trials is changed, the number of experimental outcomes also changes.




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<u>Given that:</u>

ΔUVW,

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