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Andrei [34K]
1 year ago
8

X. Y

Mathematics
1 answer:
Dovator [93]1 year ago
4 0

Answer:

Slope:  5

Y-intercept:  0

Equation:  y=5x

Step-by-step explanation:

Lets look for an equation of the format y = mx + b, where m is the slope and b is the y-intercept (the value of y when x = 0).

Slope is defined as the "Rise/Run" of the line.  The change in y(the rise) for a change in x(the run).  This can be calculated by taking any two of the given data points.  I'll pick (1,5) and (5,25):

Rise = (25 - 5) = 20

Run = (5 - 1) = 4

The Rise/Run, or slope, m, is (20/4) or 5.

<u>The equation becomes y = 5x + b.</u>

To find b, the y-intercept, enter <u>any</u> of the points into the equation and solve for b:

y = 5x + b

y = 5x + b   for (4,20)

20 = 5*(4) + b

b = 0

The line goes through the origin at x = 0 (0,0).

The equation is y = 5x + 0 or just y = 5x.

Slope:  5

Y-intercept:  0

Equation:  y=5x

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Answer:

(a) The distribution of X=\sum\limits^{n}_{i=1}{X_{i}} is a Binomial distribution.

(b) The sampling distribution of the sample mean will be approximately normal.

(c) The value of P(\bar X>0.50) is 0.50.

Step-by-step explanation:

It is provided that random variables X_{i} are independent and identically distributed Bernoulli random variables with <em>p</em> = 0.50.

The random sample selected is of size, <em>n</em> = 100.

(a)

Theorem:

Let X_{1},\ X_{2},\ X_{3},...\ X_{n} be independent Bernoulli random variables, each with parameter <em>p</em>, then the sum of of thee random variables, X=X_{1}+X_{2}+X_{3}...+X_{n} is a Binomial random variable with parameter <em>n</em> and <em>p</em>.

Thus, the distribution of X=\sum\limits^{n}_{i=1}{X_{i}} is a Binomial distribution.

(b)

According to the Central Limit Theorem if we have an unknown population with mean <em>μ</em> and standard deviation <em>σ</em> and appropriately huge random samples (<em>n</em> > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.  

The sample size is large, i.e. <em>n</em> = 100 > 30.

So, the sampling distribution of the sample mean will be approximately normal.

The mean of the distribution of sample mean is given by,

\mu_{\bar x}=\mu=p=0.50

And the standard deviation of the distribution of sample mean is given by,

\sigma_{\bar x}=\sqrt{\frac{\sigma^{2}}{n}}=\sqrt{\frac{p(1-p)}{n}}=0.05

(c)

Compute the value of P(\bar X>0.50) as follows:

P(\bar X>0.50)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{0.50-0.50}{0.05})\\

                    =P(Z>0)\\=1-P(Z

*Use a <em>z</em>-table.

Thus, the value of P(\bar X>0.50) is 0.50.

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