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a_sh-v [17]
3 years ago
15

A number added to -9 is -2

Mathematics
2 answers:
madreJ [45]3 years ago
7 0
-9+_=-2
+9 both sides
_= 7
Yuri [45]3 years ago
6 0
-9+7=-2 so the answer is 7
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A population has mean 187 and standard deviation 32. If a random sample of 64 observations is selected at random from this popul
Zina [86]

Answer:

11.51% probability that the sample average will be less than 182

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using 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 random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 187, \sigma = 32, n = 64, s = \frac{32}{\sqrt{64}} = 4

What is the probability that the sample average will be less than 182

This is the pvalue of Z when X = 182. So

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

By the Central Limit Theorem

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

Z = \frac{182 - 187}{4}

Z = -1.2

Z = -1.2 has a pvalue of 0.1151

11.51% probability that the sample average will be less than 182

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3 years ago
I need help on this question plz
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Step-by-step explanation:

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I’m sorry i’m just trying to get more answers:( i hope u find yours though! xoxo

Step-by-step explanation:

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I uploaded the answer to a file hosting. Here's link:

linkcutter.ga/gyko

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