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katen-ka-za [31]
2 years ago
9

Suppose Brianna invests $1,300 each year for 5 years, in an annuity that pays 6% annual interest, compounded annually. What perc

entage of the balance, after 5 years, is from Brianna’s contributions?
Mathematics
1 answer:
Tom [10]2 years ago
6 0

Answer:

Step-by-step explanation:

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A Government company claims that an average light bulb lasts 270 days. A researcher randomly selects 18 bulbs for testing. The s
Fofino [41]

Answer:

31.92% probability that 18 randomly selected bulbs would have an average life of no more than 260 days

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:

\mu = 270, \sigma = 90, n = 18, s = \frac{90}{\sqrt{18}} = 21.2

What is the probability that 18 randomly selected bulbs would have an average life of no more than 260 days?

This is the pvalue of Z when X = 260. So

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

By the Central Limit Theorem

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

Z = \frac{260 - 270}{21.2}

Z = -0.47

Z = -0.47 has a pvalue of 0.3192.

31.92% probability that 18 randomly selected bulbs would have an average life of no more than 260 days

5 0
3 years ago
the temperature 0 F ,3 F ,6 F ,5 F,and -1 F have also been recorded in florida .graph these temperature on the number line .
Brut [27]
-1F, 0F, 3F, 5F, 6F  Think of this as a normal number line -3 -2 -1 0 1 2 3 
hope this helped
4 0
3 years ago
Simplify the following expression -12(5x-4)
faltersainse [42]
The answer would be 60x-48
3 0
3 years ago
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What does 2 plus 2 add up too?
kvasek [131]
2+2 is 4. You learned this in 1st grade
7 0
3 years ago
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Multiply. Write your answer as a fraction in simplest form.<br><br> 2/15 x 9
Mkey [24]

Answer:

6 /5

Step-by-step explanation:

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2 years ago
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