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baherus [9]
3 years ago
12

I need help to answer the above questions

Mathematics
1 answer:
Alika [10]3 years ago
4 0
1) BMI calculation

Subject       A                                  B                                   C

Age            57                                 30                                 18

Gender       Male                             Male                             Male

heigth         70 in                             72 in                             68
                   70 * 2.54 = 177.8cm    72*2.54=182.88cm      68*2.54=172.72 cm

weight        165 lb                           190 lb                           185 lb
                  165/2.2046=74.8kg      190/2.2046=86.2kg     185/2.2046=83.9kg

BMI            74.8*10,000                  86.2*10,000                83.9*10,000
                  ----------------- = 23.7      ----------------- = 25.8    ----------------- = 28.1
                     (177.8)^2                     (182.88)^2                  (172.72)^2

2) Weight status

Subject A: 18.5 ≤ 23.7 ≤ 24.9 => Considered Ideal

Subject B: 20.5 ≤ 25.8 ≤ 29.9 => Considered Overweight

Subject C: 20.5 ≤ 28.1 ≤ 29.9 => Considered Obese

Ideal weight range

Inequality

             w * 10,000                   18.5*h^2               24.9*h^2
 18.5 ≤ --------------- ≤ 24.9 =>  ------------- ≤ w ≤    -------------
                 h^2                           10,000                   10,000

Subject A: h = 177.8 cm =>

18.5*(177.8)^2           24.9*(177.8)^2
------------------- ≤ w ≤ ---------------------
     10,000                      10,000

          58.5 kg ≤ w ≤ 78.7

Subject B: h = 182.88 cm
            
18.5*(182.88)^2              24.9*(182.88)^2
-------------------    ≤ w ≤ ------------------------
     10,000                            10,000

          61.9 kg ≤ w ≤ 83.3

Subject C:

18.5*(172.72)^2           24.9*(172.72)^2
--------------------- ≤ w ≤ ---------------------
     10,000                         10,000

          55.2 kg ≤ w ≤ 74.3 kg

3) Is BMI an appropiate measure of health?

Pros:

* it is quick and easy to calculate, so it provides a prompt measure.

Cons:

* It is not accurate.

* It leaves out specific conditions of the person: genetic and fitness.

I like that it provides a standardization to compare the status of different persons based on an objective measure, but you need to include other factors: age, gender, build, fat %, activities performed, among others. So, I am in favor of using it only like a factor but you need to use other additional measures.
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-9/7

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Porphyrin is a pigment in blood protoplasm and other body fluids that is significant in body energy and storage. Let x be a rand
SOVA2 [1]

Answer:

<u>Question 1:</u>

<u>(a) </u>P(x<60) = 0.9236

<u>(b) </u>P(x>16) = 0.9564

<u>(c) </u>P(16<x<60) = 0.88

<u>(d) </u>P (x>60) = 0.0764

<u />

<u>Question 2:</u>

<u>(a) </u>P(x<3) = 0.0668

<u>(b) </u>P(x>7) = 0.0062

<u>(c) </u>P(3<x<7) = 0.927

<u />

Step-by-step explanation:

<u>Question 1:</u>

x = no. of mg of porphyrin per deciliter of blood.

μ = 40

σ = 14

(a) We need to compute P(x<60). We need to find the z-score using the normal distribution formula:

z = (x - μ)/σ

P(x<60) = P((x - μ)/σ < (60 - 40)/14)

             = P(z < 20/14)

             = P(z<1.43)

Using the normal distribution probability table we can find the value of p at z=1.43.

P(z<1.43) = 0.9236

so, P(x<60) = 0.9236

(b) P(x>16) = P(z>(16-40)/14)

                 = P(z>-1.71)

                 = 1 - P(z<-1.71)

                 = 1 - 0.0436

    P(x>16) = 0.9564

(c) P(16<x<60) = P((16-40)/14) < x < (60-40)/14)

                        = P(-1.71 < z < 1.43)

This probability can be calculated as: P(z<1.43) - P(z<-1.71)

                                         P(16<x<60) = 0.9236 - 0.0436

                                         P(16<x<60) = 0.88

(d) P(x>60) = 1 - P(x<60)

     we have calculated P(x<60) in part (a) so,

    P(x>60) = 1 - 0.9236

    P (x>60) = 0.0764

<u>Question 2:</u>                              

μ = 4.5 mm

σ = 1.0 mm

In this question, we will again compute the z-scores and then find the probability from the normal distribution table.

(a) P(x<3) = P(z<(3-4.5)/1)

                = P(z<-1.5)

     P(x<3) = 0.0668

(b) P(x>7) = 1 - P(x<7)

               = 1 - P(z<(7-4.5)/1)

               = 1 - P(z<2.5)

               = 1 - 0.9938

    P(x>7) = 0.0062

(c) P(3<x<7) = P(x<7) - P(x<3)

  we have computed both of these probabilities in parts (a) and (b) so,

    P(3<x<7) = 0.9938 - 0.0668

    P(3<x<7) = 0.927

6 0
3 years ago
Suppose that f(x) is the (continuous) probability density function for heights of American men, in inches, and suppose that f(69
anastassius [24]

Answer:

(a) 7.6%

(b) 46.2%    42.4%

Step-by-step explanation:

(a)According to the definition of Continuous probability distribution

f(x) = \frac{d}{dx}F(x)

f(x) = \frac{F(x + h) - F(x - h)}{(x + h) - (x - h)}

f(69) = \frac{F(69 + 0.2) - F(69 - 0.2)}{0.4}

⇒ 0.19 × 0.4 = F(69.2) - F(68.8)

⇒ F(69.2) - F(68.8) = 0.076

⇒ 7.6%

(b) Given F(69) = 0.5

f(x) = \frac{d}{dx}F(x)

f(x) = \frac{F(x) - F(x - h)}{x - (x - h)}

f(69) = \frac{F(69) - F(69 - 0.2)}{0.2}

⇒ 0.19 × 0.2 = F(69) - F(68.8)

⇒ F(68.8) = 0.5 - 0.038 = 0.462

⇒ 46.2%

f(x) = \frac{d}{dx}F(x)

f(x) = \frac{F(x) - F(x - h)}{x - (x - h)}

f(69) = \frac{F(69) - F(69 - 0.4)}{0.4}

⇒ 0.19 × 0.4 = F(69) - F(68.8)

⇒ F(68.8) = 0.5 - 0.076 = 0.424

⇒ 42.4%

5 0
3 years ago
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