By knowing how much Mark deposits, how much he deducts and what is his initial balance, we can know what the final balance is.
We will see that the balance at the end of the month is $2224.
We know that:
Mark deposited:
$450, $312, $125, and $432.
<u>The total deposit</u> is: $450 + $312 + $125 + $432 = $1319
The deductions are:
$205 and $123
<u>The total deductions</u> are: $205 + $123 = $328
<u>The initial balance</u> is $1233.
The final balance will be equal to the initial balance plus what Mark deposits minus what Mark deducts, this is:
Final balance = $1233 + $1319 - $328 = $2224
We can conclude that the balance at the end of the month is $2224
If you want to learn more, you can read:
brainly.com/question/19245500
Answer:
Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:
Where
and
From the central limit theorem we know that the distribution for the sample mean
is given by:
![\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})](https://tex.z-dn.net/?f=%5Cbar%20X%20%5Csim%20N%28%5Cmu%2C%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%29)
Part a
The mean is ![\mu_{\bar x }= 17](https://tex.z-dn.net/?f=%5Cmu_%7B%5Cbar%20x%20%7D%3D%2017)
Part b
And the deviation:
![\sigma_{\bar X}= \frac{5.6}{\sqrt{13}}= 1.553](https://tex.z-dn.net/?f=%5Csigma_%7B%5Cbar%20X%7D%3D%20%5Cfrac%7B5.6%7D%7B%5Csqrt%7B13%7D%7D%3D%201.553)
Step-by-step explanation:
Assuming this complete info: Suppose a random variable xx is normally distributed with μ=17 and σ=5.6. According to the Central Limit Theorem, for samples of size 13:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".
Solution to the problem
Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:
Where
and
From the central limit theorem we know that the distribution for the sample mean
is given by:
![\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})](https://tex.z-dn.net/?f=%5Cbar%20X%20%5Csim%20N%28%5Cmu%2C%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%29)
Part a
The mean is ![\mu_{\bar x }= 17](https://tex.z-dn.net/?f=%5Cmu_%7B%5Cbar%20x%20%7D%3D%2017)
Part b
And the deviation:
![\sigma_{\bar X}= \frac{5.6}{\sqrt{13}}= 1.553](https://tex.z-dn.net/?f=%5Csigma_%7B%5Cbar%20X%7D%3D%20%5Cfrac%7B5.6%7D%7B%5Csqrt%7B13%7D%7D%3D%201.553)
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