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ad-work [718]
2 years ago
11

In planning her​ retirement, Liza deposits some money at 3.5​% ​interest, with twice as much deposited at 4.5​%. Find the amount

deposited at each rate if the total annual interest income is ​$2250.
Mathematics
1 answer:
KATRIN_1 [288]2 years ago
5 0

a = amount deposited at 3.5%

b = amount deposited at 4.5%

we know that "b" is twice as much as "a", thus b = 2a.

\begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{3.5\% of a}}{\left( \cfrac{3.5}{100} \right)a}\implies 0.035a \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{4.5\% of 2a}}{\left( \cfrac{4.5}{100} \right)2a}\implies 0.045(2a)

we also know that whatever "a" amount is, their sum is 2250, thus

0.035a+0.045(2a) = 2250\implies 0.035a+0.09a=2250\implies 0.125a=2250 \\\\\\ a=\cfrac{2250}{0.125}\implies \boxed{a=18000}~\hspace{10em}\boxed{\stackrel{2a}{b=36000}}

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Answer:

0.0228 = 2.28% probability that the average score of the 64 golfers exceeded 76.

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 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 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.

In this problem, we have that:

\mu = 75, \sigma = 4, n = 64, s = \frac{4}{\sqrt{64}} = 0.5

Find the probability that the average score of the 64 golfers exceeded 76.

This is 1 subtracted by the pvalue of Z when X = 64.

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

By the Central Limit Theorem

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

Z = \frac{76 - 75}{0.5}

Z = 2

Z = 2 has a pvalue of 0.9772

1 - 0.9772 = 0.0228

0.0228 = 2.28% probability that the average score of the 64 golfers exceeded 76.

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