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Sergio [31]
3 years ago
10

Answer for a(x+b)=c;2(x+1)=9

Mathematics
1 answer:
babunello [35]3 years ago
6 0
Ax+ab=c; 2x2 = 9

Try to take it from there
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An investment website can tell what devices are used to access the site. The site managers wonder whether they should enhance th
scZoUnD [109]

Answer:

a) 0.047

b) 50% probability that the sample proportion of smart phone users is greater than 0.33.

c) 33.39% probability that the sample proportion is between 0.19 and 0.31

Step-by-step explanation:

To solve this question, we need to understand 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.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

In this question, we have that:

p = 0.33, n = 100

a) What would the standard deviation of the sampling distribution of the proportion of the smart phone users​ be?

s = \sqrt{\frac{0.33*0.67}{100}} = 0.047

b) What is the probability that the sample proportion of smart phone users is greater than 0.33?

This is 1 subtracted by the pvalue of Z when X = 0.33. So

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

By the Central Limit Theorem

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

Z = \frac{0.33 - 0.33}{0.047}

Z = 0

Z = 0 has a pvalue of 0.5

1 - 0.5 = 0.5

50% probability that the sample proportion of smart phone users is greater than 0.33.

c) What is the probability that the sample proportion is between 0.19 and 0.31​?

This is the pvalue of Z when X = 0.31 subtracted by the pvalue of Z when X = 0.19. So

X = 0.31

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

Z = \frac{0.31 - 0.33}{0.047}

Z = -0.425

Z = -0.425 has a pvalue of 0.3354

X = 0.19

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

Z = \frac{0.19 - 0.33}{0.047}

Z = -2.97

Z = -2.97 has a pvalue of 0.0015

0.3354 - 0.0015 = 0.3339

33.39% probability that the sample proportion is between 0.19 and 0.31

3 0
4 years ago
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