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Lady bird [3.3K]
2 years ago
9

What is the interquartile range of the set of data represented by the following box-and-whisker plot?

Mathematics
2 answers:
Valentin [98]2 years ago
7 0
In descriptive statistics, the interquartile range<span> (</span>IQR<span>), also called the midspread or middle 50%, or technically H-spread, is a measure of statistical dispersion, being equal to the difference between 75th and 25th percentiles... ~ wikipedia

according to Wikipedia and some quick mental math the answer would be four

four is the difference between 14 and 18 which are respectively the 25th and 75th percentiles.</span><span />
MissTica2 years ago
4 0

Answer:

4

Step-by-step explanation:

The interquartile range is the difference between the third quartile, Q3, and the first quartile, Q1.

In a box plot, Q1 makes up the left hand side of the box and Q3 makes up the right hand side of the box.  In this figure, Q1 is 14 and Q3 is 18; this makes the IQR

18-14 = 4

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Suppose that 40 percent of the drivers stopped at State Police checkpoints in Storrs on Spring Weekend show evidence of driving
lesantik [10]

Answer:

a) 0.778

b) 0.9222

c) 0.6826

d) 0.3174

e) 2 drivers

Step-by-step explanation:

Given:

Sample size, n = 5

P = 40% = 0.4

a) Probability that none of the drivers shows evidence of intoxication.

P(x=0) = ^nC_x P^x (1-P)^n^-^x

P(x=0) = ^5C_0  (0.4)^0 (1-0.4)^5^-^0

P(x=0) = ^5C_0 (0.4)^0 (0.60)^5

P(x=0) = 0.778

b) Probability that at least one of the drivers shows evidence of intoxication would be:

P(X ≥ 1) = 1 - P(X < 1)

= 1 - P(X = 0)

= 1 - ^5C_0 (0.4)^0 * (0.6)^5

= 1 - 0.0778

= 0.9222

c) The probability that at most two of the drivers show evidence of intoxication.

P(x≤2) = P(X = 0) + P(X = 1) + P(X = 2)

^5C_0  (0.4)^0  (0.6)^5 + ^5C_1  (0.4)^1  (0.6)^4 + ^5C_2  (0.4)^2  (0.6)^3

= 0.6826

d) Probability that more than two of the drivers show evidence of intoxication.

P(x>2) = 1 - P(X ≤ 2)

= 1 - [^5C_0  (0.4)^0  (0.6)^5 + ^5C_1  (0.4)^1  (0.6)^4 + ^5C_2 * (0.4)^2  (0.6)^3]

= 1 - 0.6826

= 0.3174

e) Expected number of intoxicated drivers.

To find this, use:

Sample size multiplied by sample proportion

n * p

= 5 * 0.40

= 2

Expected number of intoxicated drivers would be 2

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2 years ago
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Answer:

Think your answer is 3 1/4

8 0
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