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alexandr1967 [171]
2 years ago
7

Find the mean, range, and interquartile range of the heights of plants given

Mathematics
2 answers:
aleksandrvk [35]2 years ago
8 0

Answer:

Mean: 30.9

Range: 80

Interquartile range: 26

Step-by-step explanation:

To find the mean, you would add up all the numbers (2, 22, 4, 36, 82, 44, 65, 30, 8, and 16), and divide them by how many numbers are in the collection, which is 10. So you would divide 309 by ten.

To find the range, you find the largest number, which is 82 and subtracts it by the lowest number, 2. That equals 80.

Here is an example of how to find the interquartile range. I hope it helps:

Lower half     Median = 71    Upper half

62, 63, 64, 64, 70, 71, 72, 76, 77, 81, 81

Lower quarter                 Upper quarter

Interquartile range: 77-64=13

Q1 = 64                                        Q3=77

All credit goes to qbattiste.

Artist 52 [7]2 years ago
7 0

Answer:

Mean = 30.9

Range = 80

Interquartile Range = 36

Step-by-step explanation:

The mean of a set of numbers is the sum divided by the number of terms. In other words that means 2, 22, 4, 36, 82, 44, 65, 30, 8, 16 go over a fraction like this. There are 10 numbers here so this is our denominator.

\frac{2+ 22 + 4 + 36 + 82 + 44 + 65 + 30 + 8 + 16}{10}

\frac{309}{10} > 30.9

For the range, all we do is subtract the minimum value from the maximum which is 82 - 2. That gives us 80.

For the IQ Range....

Arrange the terms in ascending order.

2, 4, 8, 16, 22, 30, 36, 44, 65, 82

Then we need to find the median which is the average of the two middle terms.

\frac{22+30}{2} > \frac{2*11 + 30}{2} >\frac{2*11+2*15}{2} > \frac{2 * (11 + 15)}{2} > 11 + 15 > 26

So for the lower half of data 2, 4, 8, 16, 22 we have the median of 8

And for the upper half 30, 36, 44, 65, 82 we have the median of 44.

To find the IQ Range it is the difference between the first quartile (8) and the third (44)

So 44 - 8 = 36

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Step-by-step explanation:

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In an experiment, college students were given either four quarters or a $1 bill and they could either keep the money or spend it
gavmur [86]

Answer:

a) P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341

b) P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659

c)  A. A student given a $1 bill is more likely to have kept the money.

Because the probability 0.659 is atmoslt two times greater than 0.341

Step-by-step explanation:

Assuming the following table:

                                                     Purchased Gum      Kept the Money   Total

Students Given 4 Quarters              25                              14                      39

Students Given $1 Bill                       15                               29                    44

Total                                                   40                              43                     83

a. find the probability of randomly selecting a student who spent the money, given that the student was given a $1 bill.

For this case let's define the following events

B= "student was given $1 Bill"

A="The student spent the money"

For this case we want this conditional probability:

P(A|B) =\frac{P(A and B)}{P(B)}

We have that P(A)= \frac{40}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{15}{83}

And if we replace we got:

P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341

b. find the probability of randomly selecting a student who kept the money, given that the student was given a $1 bill.

For this case let's define the following events

B= "student was given $1 Bill"

A="The student kept the money"

For this case we want this conditional probability:

P(A|B) =\frac{P(A and B)}{P(B)}

We have that P(A)= \frac{43}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{29}{83}

And if we replace we got:

P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659

c. what do the preceding results suggest?

For this case the best solution is:

A. A student given a $1 bill is more likely to have kept the money.

Because the probability 0.659 is atmoslt two times greater than 0.341

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And the last blank is 240

Step-by-step explanation:

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7 0
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