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Solnce55 [7]
3 years ago
5

Solve the problem. A variable x has the possible observations shown below. Possible observations of x: -3 -1 0 1 1 2 4 4 5 Find

the z-score corresponding to an observed value of x of 2.
Mathematics
1 answer:
svp [43]3 years ago
4 0

Answer:

The z-score corresponding to an observed value of x of 2 is 0.215.

Step-by-step explanation:

We are given that a variable x has the possible observations shown below;

Possible observations of X: -3, -1, 0, 1, 1, 2, 4, 4, 5.

Firstly, we will find the mean and the standard deviation of X, i.e;

Mean of X, (\mu) = \frac{\sum X}{n}

                       =  \frac{(-3)+ (-1)+ 0+ 1+ 1+ 2+ 4+ 4+ 5}{9}  

                       =  \frac{13}{9}  = 1.44

Standard deviation of X, (\sigma) =  \sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }

                            =  \sqrt{\frac{(-3-1.44)^{2}+(-1-1.44)^{2}+......+(4-1.44)^{2}+(5-1.44)^{2} }{9-1} }

                            =  2.603

Now, the z-score corresponding to an observed value of x of 2 is given by;

                  z-score =  \frac{X-\mu}{\sigma}

                               =  \frac{2-1.44}{2.603}  = <u>0.215.</u>        

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Part (i)

We have 60 students total, and 5 didn't like any of the two subjects, so that must mean 60-5 = 55 students liked at least one subject.

<h3>Answer: 55</h3>

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Part (ii)

We have 15 who like math only, 20 who like science only, and 55 who like either (or both). Let x be the number of people who like both classes.

We can then say

15+20+x = 55

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<h3>Answer: 20</h3>

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Part (iii)

There are 15 people who like math only, and 20 who like both. Therefore, there are 15+20 = 35 people who like math (and some of these people also like science)

<h3>Answer: 35</h3>

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Part (iv)

We'll follow the same idea as the previous part. There are 20 people who like science only and 20 who like both subjects. That yields 40 people total who like science (and some of these people also like math).

<h3>Answer: 40</h3>

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Part (v)

We'll draw a rectangle to represent the entire group of 60 students. This is considered the universal set. Inside the rectangle will be two overlapping circles to represent math (M) and science (S).

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1. S(–4, –4), P(4, –2), A(6, 6) and Z(–2, 4) a) Apply the distance formula for each side to determine whether SPAZ is equilatera
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Answer:

a) SPAZ is equilateral.

b) Diagonals SA and PZ are perpendicular to each other.

c) Diagonals SA and PZ bisect each other.

Step-by-step explanation:

At first we form the triangle with the help of a graphing tool and whose result is attached below. It seems to be a paralellogram.

a) If figure is equilateral, then SP = PA = AZ = ZS:

SP = \sqrt{[4-(-4)]^{2}+[(-2)-(-4)]^{2}}

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PA = \sqrt{(6-4)^{2}+[6-(-2)]^{2}}

PA \approx  8.246

AZ =\sqrt{(-2-6)^{2}+(4-6)^{2}}

AZ \approx 8.246

ZS = \sqrt{[-4-(-2)]^{2}+(-4-4)^{2}}

ZS \approx 8.246

Therefore, SPAZ is equilateral.

b) We use the slope formula to determine the inclination of diagonals SA and PZ:

m_{SA} = \frac{6-(-4)}{6-(-4)}

m_{SA} = 1

m_{PZ} = \frac{4-(-2)}{-2-4}

m_{PZ} = -1

Since m_{SA}\cdot m_{PZ} = -1, diagonals SA and PZ are perpendicular to each other.

c) The diagonals bisect each other if and only if both have the same midpoint. Now we proceed to determine the midpoints of each diagonal:

M_{SA} = \frac{1}{2}\cdot S(x,y) + \frac{1}{2}\cdot A(x,y)

M_{SA} = \frac{1}{2}\cdot (-4,-4)+\frac{1}{2}\cdot (6,6)

M_{SA} = (-2,-2)+(3,3)

M_{SA} = (1,1)

M_{PZ} = \frac{1}{2}\cdot P(x,y) + \frac{1}{2}\cdot Z(x,y)

M_{PZ} = \frac{1}{2}\cdot (4,-2)+\frac{1}{2}\cdot (-2,4)

M_{PZ} = (2,-1)+(-1,2)

M_{PZ} = (1,1)

Then, the diagonals SA and PZ bisect each other.

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