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Harlamova29_29 [7]
2 years ago
14

Please help me I’ll mark you brainlyest.

Mathematics
1 answer:
Alenkasestr [34]2 years ago
7 0

Answer:

10 and 7

Step-by-step explanation:

I hope this helps you out with your question! Can you plz consider marking me brainliest and thanks! Plz tell me how the answer was! It feel nice to be appreciated! Comment /free for a giveaway entry of 20 thanks and 2 brainliest. Have an awesome day! Let me know if you have any questions!

                                                               -KaraYeaton18

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 Find the sum ​(33x+99​)+​(minus−44x+88​)
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A researcher reports survey results by stating that the standard error of the mean is 25 the population standard deviation is 40
bezimeni [28]

Answer:

a) A sample of 256 was used in this survey.

b) 45.14% probability that the point estimate was within ±15 of the population mean

Step-by-step explanation:

This question is solved using the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

a. How large was the sample used in this survey?

We have that s = 25, \sigma = 400. We want to find n, so:

s = \frac{\sigma}{\sqrt{n}}

25 = \frac{400}{\sqrt{n}}

25\sqrt{n} = 400

\sqrt{n} = \frac{400}{25}

\sqrt{n} = 16

(\sqrt{n})^2 = 16^2[tex][tex]n = 256

A sample of 256 was used in this survey.

b. What is the probability that the point estimate was within ±15 of the population mean?

15 is the bounds with want, 25 is the standard error. So

Z = 15/25 = 0.6 has a pvalue of 0.7257

Z = -15/25 = -0.6 has a pvalue of 0.2743

0.7257 - 0.2743 = 0.4514

45.14% probability that the point estimate was within ±15 of the population mean

3 0
3 years ago
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