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Marina86 [1]
3 years ago
8

Sally opens a bank account with $500. The account accrues 5% interest compounded annually.

Mathematics
1 answer:
Yuri [45]3 years ago
3 0
500 x 1.05^4 

it's 4 periods because the interest rate is compounded yearly. 5 percent yearly increase is equal to P x 1.05
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What number would you add to both sides of x2 + 7x = 4 to complete the square?
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An HP laser printer is advertised to print text documents at a speed of 18 ppm (pages per minute). The manufacturer tells you th
aliya0001 [1]

Answer:

0.227 = 22.7% probability that the mean printing speed of the sample is greater than 18.12 ppm.

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 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.

Mean of 17.42 ppm and a standard deviation of 3.25 ppm.

This means that \mu = 17.42, \sigma = 3.25

Sample of 12:

This means that n = 12, s = \frac{3.25}{\sqrt{12}}

Find the probability that the mean printing speed of the sample is greater than 18.12 ppm.

This is 1 subtracted by the p-value of Z when X = 18.12.

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

By the Central Limit Theorem

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

Z = \frac{18.12 - 17.42}{\frac{3.25}{\sqrt{12}}}

Z = 0.75

Z = 0.75 has a pvalue of 0.773.

1 - 0.773 = 0.227

0.227 = 22.7% probability that the mean printing speed of the sample is greater than 18.12 ppm.

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