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DerKrebs [107]
3 years ago
12

You have $550 in a savings account that earns 3% simple interest each yaer. How much will be in your account in 10 years?

Mathematics
1 answer:
bija089 [108]3 years ago
4 0
Looking at the given question, we can find several important information's and based on thm we can easily calculate the amount after 10 years.
Principal amount = $550
Rate of interest per year = 3%
Number of years the amount is to be kept = 10 years
Now we get put these information's in the formula of finding amount of money kepy in simple interest.
Amount = P(1 + rt)
Where
p = principal amount
r = rate of interest
t = Time
Then
Amount = 550[1 + (3/100) * 10)]
             = 550 [1 + 3/10]
             = 550 * (13/10)
             =  55 * 13
             = 715 dollars
So the total amount in my account will be $715
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Answer:

0.5034 = 50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

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 can be 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 random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean

In this problem, we have that:

\mu = 119, \sigma = 14, n = 74, s = \frac{14}{\sqrt{74}} = 1.6275

What is the probability that the sample mean would differ from the true mean by less than 1.11 months?

This is the pvalue of Z when X = 119 + 1.1 = 120.1 subtracted by the pvalue of Z when X = 119 - 1.1 = 117.9. So

X = 120.1

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

By the Central Limit Theorem

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

Z = \frac{120.1 - 119}{1.6275}

Z = 0.68

Z = 0.68 has a pvalue of 0.7517

X = 117.9

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

Z = \frac{117.9 - 119}{1.6275}

Z = -0.68

Z = -0.68 has a pvalue of 0.2483

0.7517 - 0.2483 = 0.5034

0.5034 = 50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

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Answer:

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