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Doss [256]
3 years ago
7

What is the APY of a savings account with an APR of 3.9742% compounded monthly? Answer to the four decimal place

Mathematics
2 answers:
Maurinko [17]3 years ago
6 0

Answer:

The APY of the saving account is 4.0474%

Step-by-step explanation:

We know the formula for APY which is given by

APY=(1+\frac{r}{n} )^n-1

here, r= interset rate =  3.9742% = 0.039742

n = compounding cycles = 12

On plugging these values in the above formula, we get

APY=(1+\frac{0.039742}{12} )^{12}-1

On simplifying this we get

APY =0.04047395=4.0474%

LUCKY_DIMON [66]3 years ago
3 0

Answer:

APY = 4.0474%

Step-by-step explanation:

Given is the Annual Percentage Rate (APR) = 3.9742% compounded monthly.

Suppose Principal amount, P = $1.

time, t = 1 year.

interest rate, r = 3.9742% = 0.039742

period of compounding, n = 12 (for monthly).

Future value = P * (1 + r/n)^(nt)

FV = 1 * (1 + 0.039742/12)^(1*12) = 1.040473955

Annual Percentage Growth = (FV/P)*100 = (1.040473955 / 1) * 100 = 4.0473955%

Hence, Annual Percentage Yield (APY) = 4.0474%

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Answer:

The "probability that a given score is less than negative 0.84" is  \\ P(z.

Step-by-step explanation:

From the question, we have:

  • The random variable is <em>normally distributed</em> according to a <em>standard normal distribution</em>, that is, a normal distribution with \\ \mu = 0 and \\ \sigma = 1.
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A z-score is a standardized value, i.e., one that we can obtain using the next formula:

\\ z = \frac{x - \mu}{\sigma} [1]

Where

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  • And we already know that \\ \mu and \\ \sigma are the mean and the standard deviation, respectively, of the <em>normal distribution</em>.

A <em>z-score</em> represents the <em>distance</em> from \\ \mu in <em>standard deviations</em> units. When the value for z is <em>negative</em>, it "tells us" that the raw score is <em>below</em> \\ \mu. Conversely, when the z-score is <em>positive</em>, the standardized raw score, <em>x</em>, is <em>above</em> the mean, \\ \mu.

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We already know that \\ z = -0.84 or that the standardized value for a raw score, <em>x</em>, is <em>below</em> \\ \mu in <em>0.84 standard deviations</em>.

The values for probabilities of the <em>standard normal distribution</em> are tabulated in the <em>standard normal table, </em>which is available in Statistics books or on the Internet and is generally in <em>cumulative probabilities</em> from <em>negative infinity</em>, - \\ \infty, to the z-score of interest.

Well, to solve the question, we need to consult the <em>standard normal table </em>for \\ z = -0.84. For this:

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This represent the area under the <em>standard normal distribution</em>, \\ N(0,1), at the <em>left</em> of <em>z = -0.84</em>.

To "draw a sketch of the region", we need to draw a normal distribution <em>(symmetrical bell-shaped distribution)</em>, with mean that equals 0 at the middle of the distribution, \\ \mu = 0, and a standard deviation that equals 1, \\ \sigma = 1.

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Find the place where z = -0.84 (i.e, below the mean and near to negative one standard deviation, \\ -\sigma, from it). All the area to the left of this value must be shaded because it represents \\ P(z and that is it.

The below graph shows the shaded area (in blue) for \\ P(z for \\ N(0,1).

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