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nikitadnepr [17]
1 year 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]1 year 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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<span>I think the answer is false. </span>
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Answer:

we have to maximize the following equation:

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where:

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C = number of model C bicycles produced

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using solver, the optimal solution is: 745A + 1006B = $83,825

using slack variables:

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slack variable tableau:

A         B         C           S1          S2          S3           Z            B

2         2.5       3           1             0            0            0            4006

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Answer:

a) ME=2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.0423

b) 0.413 - 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.371

Step-by-step explanation:

1) Data given and notation  

n=900 represent the random sample taken    

X=372 represent the students were pursuing liberal arts degrees

\hat p=\frac{372}{900}=0.413 estimated proportion of students were pursuing liberal arts degrees

\alpha=0.01 represent the significance level

z would represent the statistic (variable of interest)    

p_v represent the p value (variable of interest)    

p= population proportion of students were pursuing liberal arts degrees

2) Solution to the problem

The confidence interval would be given by this formula

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ME=2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.0423

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Answer:

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