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fomenos
3 years ago
8

The two box-and-whisker plots below represent the distribution of shoe sizes for a group of 40 two-year-old boys and a group of

80 three-year-old boys.
What is the difference of the interquartile ranges of the box-and-whisker plots?

0
0.5
1
1.5

Mathematics
1 answer:
FromTheMoon [43]3 years ago
7 0
<h3>Answer: Choice B)  0.5</h3>

=============================================================

Explanation:

For now, focus solely on the orange boxplot. The first quartile Q1 is the left edge of the box, which is at 3.5; while the value of Q3 is 7.5 (right edge of the box). The interquarile range (IQR) is ...

IQR = Q3 - Q1

IQR = 7.5 - 3.5

IQR = 4

This is basically the width of the box. Ignore the whiskers when it comes to the IQR.

Let A = 4 as we'll use it later.

---------------------------------

Find the IQR for the blue box plot

Q1 = 6 = left edge of the blue box

Q3 = 9.5 = right edge of the blue box

IQR = Q3 - Q1

IQR = 9.5 - 6

IQR = 3.5

Let B = 3.5 as we'll use it later

---------------------------------

Subtract the values of A and B to find the difference of the IQR values

A - B = 4 - 3.5 = 0.5

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Kisachek [45]

Answer:

(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b) The probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c) The probability that at most five must be selected to find four that are not seconds is 0.9453.

Step-by-step explanation:

Let <em>X</em> = number of seconds in the batch.

The probability of the random variable <em>X</em> is, <em>p</em> = 0.31.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X</em> is:

P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...

(a)

Compute the probability that only one goblet is a second among six randomly selected goblets as follows:

P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888

Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b)

Compute the probability that at least two goblet is a second among six randomly selected goblets as follows:

P (X ≥ 2) = 1 - P (X < 2)

              =1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776

Thus, the probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c)

If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.

P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)

                            ={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453

Thus, the probability that at most five must be selected to find four that are not seconds is 0.9453.

8 0
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Salvador is playing a game where the product must be greater than 2. Which two cards result in a product greater than 2? 1/2, 1
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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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4(x+9)  

Solution

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