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nikitadnepr [17]
2 years ago
7

The data represents the body mass index​ (BMI) values for 20 females. Construct a frequency distribution beginning with a lower

class limit of 15.0 and use a class width of 6.0. Does the frequency distribution appear to be roughly a normal distribution?
17.7
29.4
19.2
27.5
33.5
25.6
22.1
44.9
26.5
18.3
22.4
32.4
24.9
28.6
37.7
26.1
21.8
21.2
30.7
21.4
Body Mass Index Frequency
15.0 dash 20.9 nothing
21.0 dash 26.9 nothing
27.0 dash 32.9 nothing
Body Mass Index Frequency
33.0 dash 38.9 nothing
39.0 dash 44.9 nothing
Mathematics
1 answer:
Papessa [141]2 years ago
3 0

Answer:

Given:

Body mass index values:

17.7

29.4

19.2

27.5

33.5

25.6

22.1

44.9

26.5

18.3

22.4

32.4

24.9

28.6

37.7

26.1

21.8

21.2

30.7

21.4

Constructing a frequency distribution beginning with a lower class limit of 15.0 and use a class width of 6.0.

we have:

Body Mass Index____ Frequency

15.0 - 20.9__________3( values of 17.7, 18.3, & 19.2 are within this range)

21.0 to 26.9__________8 values are within this range)

27.0 - 32.9____________ 5 values

33.0 - 38.9____________ 2 values

39.0 - 44.9 _____________2 values

The frequency distribution is not a normal distribution. Here, although the frequencies start from the lowest,  increases afterwards and then a decrease is recorded again, it is not normally distributed because it is not symmetric.

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A construction company is considering submitting bids for contracts of three different projects. The company estimates that it h
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Answer:

a.P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}\\

b. E(x) = 0.3

c. S(x)=0.5196

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Step-by-step explanation:

The probability that the company won x bids follows a binomial distribution because we have n identical and independent experiments with a probability p of success and (1-p) of fail.

So, the PMF of X is equal to:

P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}\\

Where p is 0.1 and it is the chance of winning. Additionally, n is 3 and it is the number of bids. So the PMF of X is:

P(x)=\frac{3!}{x!(3-x)!}*0.1^{x}*(1-0.1)^{n-x}\\

For binomial distribution:

E(x)=np\\S(x)=\sqrt{np(1-p)}

Therefore, the company can expect to win 0.3 bids and it is calculated as:

E(x) = np = 3*0.1 = 0.3

Additionally, the standard deviation of the number of bids won is:

S(x)=\sqrt{np(1-p)}=\sqrt{3(0.1)(1-0.1)}=0.5196

Finally, the probability to won 1, 2 or 3 bids is equal to:

P(1)=\frac{3!}{1!(3-1)!}*0.1^{1}*(1-0.1)^{3-1}=0.243\\P(2)=\frac{3!}{2!(3-2)!}*0.1^{2}*(1-0.1)^{3-2}=0.027\\P(3)=\frac{3!}{3!(3-3)!}*0.1^{3}*(1-0.1)^{3-3}=0.001

So, the expected profit for the company is equal to:

E=-10,000+50,000(0.243)+100,000(0.027)+150,000(0.001)\\E=5,000

Because there is a probability of 0.243 to win one bid and it will produce 50,000 of income, there is a probability of 0.027 to win 2 bids and it will produce 100,000 of income and there is a probability of 0.001 to win 3 bids and it will produce 150,000 of income.

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