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statuscvo [17]
3 years ago
9

Anne is comparing savings accounts. One account has an interest rate of 1.2 percent compounded yearly, and one account has an in

terest rate of 1.2 percent compounded monthly. Which account will earn more money in interest?
Mathematics
1 answer:
Varvara68 [4.7K]3 years ago
5 0
In short and without much fuss

let's say Anne puts "x" amount in the account at 1.2% rate annually, that means after 1 year, she will have "x" + 1.2% of "x", or 1.012x to be exact.

the 1.2% rate, kicks in as the period of a year is met.

now, what if Anne puts it in the monthly compounded type?  that means, the compounding period is a month, so after 1 month, she has 1.2% extra, or 1.012x, and after 2 months, she will have 1.2% extra of 1.012x, or 1.012144x, and after 3 months, she will have 1.2% extra of 1.0121x, or 1.012145728x and so on.

anyhow, the shorter the compounding period, the more the 1.2% kicks in, the more accumulation in the account.
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The mean of a certain set of measurements is 27 with a standard deviation of 14. The distribution of the
Talja [164]

Answer:

Exactly 16%.

Step-by-step explanation:

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.

The mean of a certain set of measurements is 27 with a standard deviation of 14.

This means that \mu = 27, \sigma = 14

The proportion of measurements that is less  than 13 is

This is the p-value of Z when X = 13, so:

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

Z = \frac{13 - 27}{14}

Z = -1

Z = -1 has a p-value of 0.16, and thus, the probability is: Exactly 16%.

8 0
2 years ago
Let X1 and X2 be independent random variables with mean μand variance σ².
My name is Ann [436]

Answer:

a) E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu

So then we conclude that \hat \theta_1 is an unbiased estimator of \mu

E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu

So then we conclude that \hat \theta_2 is an unbiased estimator of \mu

b) Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2}

Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8}

Step-by-step explanation:

For this case we know that we have two random variables:

X_1 , X_2 both with mean \mu = \mu and variance \sigma^2

And we define the following estimators:

\hat \theta_1 = \frac{X_1 + X_2}{2}

\hat \theta_2 = \frac{X_1 + 3X_2}{4}

Part a

In order to see if both estimators are unbiased we need to proof if the expected value of the estimators are equal to the real value of the parameter:

E(\hat \theta_i) = \mu , i = 1,2

So let's find the expected values for each estimator:

E(\hat \theta_1) = E(\frac{X_1 +X_2}{2})

Using properties of expected value we have this:

E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu

So then we conclude that \hat \theta_1 is an unbiased estimator of \mu

For the second estimator we have:

E(\hat \theta_2) = E(\frac{X_1 + 3X_2}{4})

Using properties of expected value we have this:

E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu

So then we conclude that \hat \theta_2 is an unbiased estimator of \mu

Part b

For the variance we need to remember this property: If a is a constant and X a random variable then:

Var(aX) = a^2 Var(X)

For the first estimator we have:

Var(\hat \theta_1) = Var(\frac{X_1 +X_2}{2})

Var(\hat \theta_1) =\frac{1}{4} Var(X_1 +X_2)=\frac{1}{4} [Var(X_1) + Var(X_2) + 2 Cov (X_1 , X_2)]

Since both random variables are independent we know that Cov(X_1, X_2 ) = 0 so then we have:

Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2}

For the second estimator we have:

Var(\hat \theta_2) = Var(\frac{X_1 +3X_2}{4})

Var(\hat \theta_2) =\frac{1}{16} Var(X_1 +3X_2)=\frac{1}{4} [Var(X_1) + Var(3X_2) + 2 Cov (X_1 , 3X_2)]

Since both random variables are independent we know that Cov(X_1, X_2 ) = 0 so then we have:

Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8}

7 0
3 years ago
Find the sum of the first six terms of the sequence for which a2=0.7 and a3=0.49.
alina1380 [7]
A1 = 0.91
a6 = -0.14
sum = (0.91+-0.14) /2
sum = 2.31
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3 years ago
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Sphinxa [80]
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Scorpion4ik [409]

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it is an adjacent/linear pair

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