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3 years ago

8

The distribution of SAT scores of all college-bound seniors taking the SAT in 2014 was approximately normal with a mean of 14971

497 1497 1497 and standard deviation of 322 322 322 322 . Let X=X= X= X, equals the score of a randomly selected tester from this group. Find P(X<1200)P(X<1200) P(X<1200) P, (, X, is less than, 1200, ).

2 answers:

RideAnS [48]3 years ago

8 0

Answer:

P(X<1200) is 0.8212

Step-by-step explanation:

Test statistic (z) = (X - mean)/sd

X is score of a tester = 1200

mean = 1497

sd = 322

z = (1200 - 1497)/322 = -297/322 = -0.92

The cumulative area of the test statistic is the probability that X<1200. The cumulative area is 0.8212.

Therefore, P(X<1200) = 0.8212

kvasek [131]3 years ago

4 0

Given Information:

Mean = μ = 1497

Standard deviation = σ = 322

test value = x = 1200

Required Information:

P(x < 1200) = ?

Answer:

P(x < 1200) = 0.17879

Explanation:

First we will find the z-score

P(x < X) = P(z < (x - μ)/σ)

P(x < 1200) = P(z < (1200 - 1497)/322)

P(x < 1200) = P(z < -0.92)

The z-score corresponding to z < -0.92 from z-table is given by

P(z < -0.92) = 0.17879

Therefore, the probability that SAT the test score will be less than 1200 is 0.17879.

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b) Figure attached

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Step-by-step explanation:

We assume that th data is this one:

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y: 38, 43, 29, 32, 26, 33, 19, 27, 23, 14, 19, 21.

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For this case we need to calculate the slope with the following formula:

$m=\frac{S_{xy}}{S_{xx}}$

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$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}$

So we can find the sums like this:

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$\sum_{i=1}^n x^2_i =30^2+30^2+30^2+50^2+50^2+50^2+70^2+70^2+70^2+90^2+90^2+90^2=49200$

$\sum_{i=1}^n y^2_i =38^2+43^2+29^2+32^2+26^2+33^2+19^2+27^2+23^2+14^2+19^2+21^2=9540$

$\sum_{i=1}^n x_i y_i =30*38+30*43+30*29+50*32+50*26+50*33+70*19+70*27+70*23+90*14+90*19+90*21=17540$

With these we can find the sums:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=49200-\frac{720^2}{12}=6000$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}=17540-\frac{720*324}{12}{12}=-1900$

And the slope would be:

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Nowe we can find the means for x and y like this:

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$\bar y= \frac{\sum y_i}{n}=\frac{324}{12}=27$

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For this case we can use excel and we got the figure attached as the result.

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In oder to calculate S^2 we need to calculate the MSE, or the mean square error. And is given by this formula:

$MSE=\frac{SSE}{df_{E}}$

The degred of freedom for the error are given by:

$df_{E}=n-2=12-2=10$

We can calculate:

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And now we can calculate the sum of squares for the regression given by:

$SSR=\frac{S^2_{xy}}{S_{xx}}=\frac{(-1900)^2}{6000}=601.67$

We have that SST= SSR+SSE, and then SSE=SST-SSR= 792-601.67=190.33[/tex]

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$S^2=\hat \sigma^2=MSE=\frac{190.33}{10}=19.03$

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