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-BARSIC- [3]
3 years ago
9

If your income of $1500 increased by 15%, find the amount it has increased by

Mathematics
2 answers:
Sedaia [141]3 years ago
7 0

Answer:

$1725

Step-by-step explanation:

Increment=15% of $1500

=15/100×$1500

=15×15

=$225 increased

Present amount=$1500+$225

=$1725

mamaluj [8]3 years ago
3 0

Answer:

it has increased by $225

Step-by-step explanation:

=1500*15/100

=225

total income is $1725 and it has increased by 225

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Suppose 46% of politicians are lawyers. If a random sample of size 662 is selected, what is the probability that the proportion
Svet_ta [14]

Answer:

0.9606 = 96.06% probability that the proportion of politicians who are lawyers will differ from the total politicians proportion by less than 4%

Step-by-step explanation:

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

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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}}

Suppose 46% of politicians are lawyers.

This means that p = 0.46

Sample of size 662

This means that n = 662

Mean and standard deviation:

\mu = p = 0.46

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.46*0.54}{662}} = 0.0194

What is the probability that the proportion of politicians who are lawyers will differ from the total politicians proportion by less than 4%?

p-value of Z when X = 0.46 + 0.04 = 0.5 subtracted by the p-value of Z when X = 0.46 - 0.04 = 0.42. So

X = 0.5

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

By the Central Limit Theorem

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

Z = \frac{0.5 - 0.46}{0.0194}

Z = 2.06

Z = 2.06 has a p-value of 0.9803

X = 0.42

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

Z = \frac{0.42 - 0.46}{0.0194}

Z = -2.06

Z = -2.06 has a p-value of 0.0197

0.9803 - 0.0197 = 0.9606

0.9606 = 96.06% probability that the proportion of politicians who are lawyers will differ from the total politicians proportion by less than 4%

8 0
3 years ago
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