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1 month ago

Make a box-and-whisker plot for the data. What is the lower quartile value? 42 48 43 65 58 47 60 56 52 64 51 66 56 62

Mathematics

2 answers:

Zina [12.3K]1 month ago

3 0

Initially, we organize our fourteen data points:

42 43 47 48 51 52 56 56 58 60 62 64 65 66

With fourteen points, the median will be the average of the 7th and 8th, while the lower quartile corresponds to the fourth and the upper quartile is the eleventh.

The median value is 56. The upper quartile is 62, while the lower quartile is, wait for it,

result: 48

The minimum observation is 42, and the maximum is 66. Therefore, our box-and-whisker plot will show a left whisker at 42, a left box edge at 48, median line at 56, right box edge at 62, and a right whisker at 66.

Zina [12.3K]1 month ago

3 0

Conclusion: 48

Detailed process:

The data provided: 42 48 43 65 58 47 60 56 52 64 51 66 56 62

Arrange in ascending order: 42, 43, 47, 48, 51, 52, 56, 56, 58, 60, 62, 64, 65, 66

As the lower quartile represents the median of the lower half.

the lower half of the dataset: 42, 43, 47, 48, 51, 52, 56

Median of the lower half = 48

Hence, the lower quartile = 48

The upper quartile denotes the median of the upper half.

Upper half of the dataset: 56, 58, 60, 62, 64, 65, 66

Median of the upper half = 62

Thus, the upper quartile = 62

Given that the number of data points is 14

The median constitutes the average of the two middlemost values.

Median of the dataset: $\dfrac{56+56}{2}=56$

Least value = 42

Highest value = 66

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Complete Question

The complete question appears in the first uploaded image

Answer:

a) Yes, selecting 15 corresponds to a binomial experiment

d) $P(r = 15) = 3.2768 *10^{-11}$

Step-by-step explanation:

Regarding question a:

For an experiment to qualify as binomial

the trials have to be independent

each trial must yield one of two possible outcomes

Given that the selection of 15 individuals is random, we ascertain that the trials are independent and the outcomes are “either the individual saves for retirement or does not save for retirement.”

Therefore, we conclude that the selection of 15 people at random is indeed a binomial experiment.

In question b:

The probability that all selected adults do not save for retirement is mathematically modeled as

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Here C signifies combination

r = 15 implies all selected adults

n refers to the population size equating to 15

From the problem, p = 0.20

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=> $q = 1 - p$

Thus

$P(r = 15) = ^{15}C_{15} * p^{15} * q^{15-15}$

$P(r = 15) = 3.2768 *10^{-11}$ Regarding question c:

The probability that exactly five of the selected adults do not save for retirement is mathematically modeled as

$P(r = 5) = ^{15} C_5 * (0.20)^5 * (0.80)^{15}$

$P(r = 5) = 0.1032$

In relation to question d:

The probability that at least one of the selected adults opts not to save for retirement can be mathematically expressed as

$P(r \ge 1 ) = 1 - P (r = 0 )$

$P(r \ge 1 ) = 1 - [ ^{15} C _ 0 * (0.20)^{0} * (0.80 )^{15}]$

$P(r \ge 1 ) = 1 - 0.0352$

$P(r \ge 1 ) = 0.9648$

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