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stellarik [79]
3 years ago
10

Identify the class width, class midpoints, and class boundaries for the given frequency distribution.Daily Low Tempurature Frequ

encey Daily Low Tempurature full data set32-24 1 44-46 735-37 3 47-49 738-40 5 50-52 141-43 11
Mathematics
1 answer:
Natali [406]3 years ago
8 0

Answer:

I . class width is 3

ii. Class midpoints are 33,36,39,42,45,48,51.

iii. Class boundaries are: 31.5-34.5, 34.5-37.5, 37.5-40.5, 40.5-43.5, 43.5-46.5, 46.5-49.5, 49.5-52.5

Step-by-step explanation:

I. The class width is the difference between upper class boundary and lower class boundary.

ii. Class midpoints is the summation of upper and lower class limits, divided by 2.

iii. Class boundaries is the subtraction of 0.5 from the lower classe limits and the addition of 0.5 to upper class limits.

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Answer:

a) 0.152 = 15.2% probability that this person is a female who engages in physical exercise activities during the lunch hour.

b) 0.248 = 24.8% probability that this person is a female who does not engage in physical exercise activities during the lunch hour.

Step-by-step explanation:

Question a:

20% of employees engage in physical exercise.

This 20% is composed by:

8% of 60%(males)

x% of 100 - 60 = 40%(females).

Then, x is given by:

0.08*0.6 + 0.4x = 0.2

0.4x = 0.2 - 0.08*0.6

x = \frac{0.2 - 0.08*0.6}{0.4}

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Probability of being a female who engages in exercise:

40% are female, 38% of 40% engage in exercise. So

0.38*0.4 = 0.152

0.152 = 15.2% probability that this person is a female who engages in physical exercise activities during the lunch hour.

B. If we choose an employee at random from this corporation,what is the probability that this person is a female who does not engage in physical exercise activities during the lunch hour?

40% are female, 100% - 38% = 62% of 40% do not engage in exercise. So

0.62*0.4 = 0.248

0.248 = 24.8% probability that this person is a female who does not engage in physical exercise activities during the lunch hour.

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