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mario62 [17]
3 years ago
6

Obtaining a measure of intelligence from a group of college students would likely yield a somewhat normal distribution (that is,

there shouldn't be any extreme outliers). What would be the best measure of central tendency to use in this example?A. Median.B. Any of the three measures of central tendencC. MeanD. Mode
Mathematics
1 answer:
ki77a [65]3 years ago
7 0

Answer:

C. Mean

Step-by-step explanation:

We have been given that obtaining a measure of intelligence from a group of college students would likely yield a somewhat normal distribution (that is, there shouldn't be any extreme outliers).

We know that median is best measure of central tendency with extreme outliers, while mean is the best measure of central tendency when the data is normally distributed.

Mode is used when data are measured in a nominal scale.

Since the measure of intelligence from a group of college students yield a somewhat normal distribution, therefore, mean will be the best measure of central tendency.  

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Suppose you pay a dollar to roll two dice. if you roll 5 or a 6 you Get your dollar back +2 more just like it the goal will be t
LiRa [457]

Answer:

(a)$67

(b)You are expected to win 56 Times

(c)You are expected to lose 44 Times

Step-by-step explanation:

The sample space for the event of rolling two dice is presented below

(1,1), (2,1), (3,1), (4,1), (5,1), (6,1)\\(1,2), (2,2), (3,2), (4,2), (5,2), (6,2)\\(1,3), (2,3), (3,3), (4,3), (5,3), (6,3)\\(1,4), (2,4), (3,4), (4,4), (5,4), (6,4)\\(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)\\(1,6), (2,6), (3,6), (4,6), (5,6), (6,6)

Total number of outcomes =36

The event of rolling a 5 or a 6 are:

(5,1), (6,1)\\ (5,2), (6,2)\\( (5,3), (6,3)\\ (5,4), (6,4)\\(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)\\(1,6), (2,6), (3,6), (4,6), (5,6), (6,6)

Number of outcomes =20

Therefore:

P(rolling a 5 or a 6)  =\dfrac{20}{36}

The probability distribution of this event is given as follows.

\left|\begin{array}{c|c|c}$Amount Won(x)&-\$1&\$2\\&\\P(x)&\dfrac{16}{36}&\dfrac{20}{36}\end{array}\right|

First, we determine the expected Value of this event.

Expected Value

=(-\$1\times \frac{16}{36})+ (\$2\times \frac{20}{36})\\=\$0.67

Therefore, if the game is played 100 times,

Expected Profit =$0.67 X 100 =$67

If you play the game 100 times, you can expect to win $67.

(b)

Probability of Winning  =\dfrac{20}{36}

If the game is played 100 times

Number of times expected to win

=\dfrac{20}{36} \times 100\\=56$ times

Therefore, number of times expected to loose

= 100-56

=44 times

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4 years ago
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astraxan [27]

4³ ⋅ (100 ÷25) − 5^0

[parenthesis goes first, then exponents]

[any coefficient with zero exponent is equal to 1]

64 · 4 - 1

[multiplication, then subtraction]

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The diagram shows the number of quarts of blood the human heart pumps per minute. A drawing shows an anatomy of human heart. Lab
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Answer:

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Substituting x = 3 into the first expression:

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The difference of twice a number g and 10 is 24
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The number g is 7
24-10 = 14
14/2=6
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