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iris [78.8K]
3 years ago
14

Which measure would be the most appropriate to describe the center of the data in the histogram below? A-mean B-range C-median D

-interquartile range

Mathematics
2 answers:
Alex787 [66]3 years ago
7 0
I would probably choose the C-median.

The other measures give more weight to the lower temperatures, which seems to me to ignore the fact that the most probable high temperature is above the average.
S_A_V [24]3 years ago
5 0

The center of data of histogram is the median and/or mean of the data.The center of a distribution is the middle of any data.

when the data is skewed, the median is more appropriate to use as the measure of a typical value.

Mean is used as the measure of center  when the data is  symmetric.

In the given Histogram the data is skewed and so Median describe the center of the data in the histogram.

Option C- median is the right answer.

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You are working in a primary care office. Flu season is starting. For the sake of public health, it is critical to diagnose peop
Aleksandr-060686 [28]

Answer:

E. 0.11

Step-by-step explanation:

We have these following probabilities:

A 10% probability that a person has the flu.

A 90% probability that a person does not have the flu, just a cold.

If a person has the flu, a 99% probability of having a runny nose.

If a person just has a cold, a 90% probability of having a runny nose.

This can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

P = \frac{P(B).P(A/B)}{P(A)}

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

In this problem, we have that:

What is the probability that a person has the flu, given that she has a runny nose?

P(B) is the probability that a person has the flu. So P(B) = 0.1.

P(A/B) is the probability that a person has a runny nose, given that she has the flu. So P(A/B) = 0.99.

P(A) is the probability that a person has a runny nose. It is 0.99 of 0.1 and 0.90 of 0.90. So

P(A) = 0.99*0.1 + 0.9*0.9 = 0.909

What is the probability that this person has the flu?

P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.1*0.99}{0.909} = 0.1089 = 0.11

The correct answer is:

E. 0.11

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