Data from the article “The Osteological Paradox: Problems inferring Prehistoric Health from Skeletal Samples” (Current Anthropology (1992):3

Question

Data from the article “The Osteological Paradox: Problems inferring Prehistoric Health from Skeletal Samples” (Current Anthropology (1992):343-370) suggests that a reasonable model for the distribution of heights of 5-year old children (in centimeters) is N(100, 62) . Let the letter X represent the variable “height of 5-year old”, and use this information to answer the following. Use 4 decimal places unless otherwise indicated.

(a) P(X > 89.2) =
(b) P(X < 109.78) =
(c) P(97 < X < 106) =
(d) P(X < 85.6 or X > 111.4) =
(e) P(X > 103) =
(f) P(X < 98.2) =
(g) P(100 < X < 124)=
(h) The middle 80% of all heights of 5 year old children fall between and . (Use 2 decimal places.)

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Margaret 2 months 2021-10-09T03:10:33+00:00 1 Answer 0 views 0

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    2021-10-09T03:12:02+00:00

    Answer:

    (a) P (X < 109.78) = 0.9484.

    (b) P (X < 109.78) = 0.9484.

    (c) P (97 < X < 106) = 0.5328.

    (d) P (X < 85.6 or X > 111.4) = 0.0369.

    (e) P (X > 103) = 0.3085.

    (f) P (X < 98.2) = 0.3821.

    (g) P (100 < X < 124) = 0.5000.

    (h) The middle 80% of all heights of 5 year old children fall between 92.31 and 107.70.

    Step-by-step explanation:

    It is provided that X follows a Normal distribution with mean, μ = 100 and standard deviation, σ = 6.

    (a)

    Compute the value of P (X > 89.2) as follows:

    P (X>89.2)=P(\frac{X-\mu}{\sigma}>\frac{89.2-100}{6})\\=P(Z>-1.80)\\=P(Z<1.8)\\=0.9641

    Thus, the value of P (X > 89.2) is 0.9641.

    (b)

    Compute the value of P (X < 109.78) as follows:

    P (X<109.78)=P(\frac{X-\mu}{\sigma}<\frac{109.78-100}{6})\\=P(Z<1.63)\\=0.9484

    Thus, the value of P (X < 109.78) is 0.9484.

    (c)

    Compute the value of P (97 < X < 106) as follows;

    P (97 < X < 106) = P (X < 106) – P (X < 97)

                              =P(\frac{X-\mu}{\sigma}<\frac{106-100}{6})+P(\frac{X-\mu}{\sigma}<\frac{97-100}{6})\\=P(Z<1)-P(Z<-0.5)\\=0.8413-0.3085\\=0.5328

    Thus, the value of P (97 < X < 106) is 0.5328.

    (d)

    Compute the value of P (X < 85.6 or X > 111.4) as follows;

    P (X < 85.6 or X > 111.4) = P (X < 85.6) + P (X > 111.4)

                                           =P(\frac{X-\mu}{\sigma}<\frac{85.6-100}{6})+P(\frac{X-\mu}{\sigma}>\frac{111.4-100}{6})\\=P(Z<-2.4)-P(Z>1.9)\\=0.0082+0.0287\\=0.0369

    Thus, the value of P (X < 85.6 or X > 111.4) is 0.0369.

    (e)

    Compute the value of P (X > 103) as follows:

    P (X>103)=P(\frac{X-\mu}{\sigma}>\frac{103-100}{6})\\=P(Z>0.50)\\=14-P(Z<0.50)\\=1-0.6915\\=0.3085

    Thus, the value of P (X > 103) is 0.3085.

    (f)

    Compute the value of P (X < 98.2) as follows:

    P (X<98.2)=P(\frac{X-\mu}{\sigma}<\frac{98.2-100}{6})\\=P(Z<-0.30)\\=1-P(Z<0.30)\\=1-0.6179\\=0.3821

    Thus, the value of P (X < 98.2) is 0.3821.

    (g)

    Compute the value of P (100 < X < 124) as follows;

    P (100< X < 124) = P (X < 124) – P (X < 100)

                              =P(\frac{X-\mu}{\sigma}<\frac{124-100}{6})+P(\frac{X-\mu}{\sigma}<\frac{100-100}{6})\\=P(Z<4)-P(Z<0)\\=1-0.50\\=0.50

    Thus, the value of P (100 < X < 124) is 0.5000.

    (h)

    Compute the value of x₁ and x₂ as follows if P (x₁ < X < x₂) = 0.80 as follows:

    P(X_{1}<X<x_{2})=P(X<x_{2})-P(X<x_{1})\\0.80=P(Z<z)-P(Z<-z)\\0.80=P(Z<z)-[1-P(Z<z)]\\0.80=2P(Z<z)-1\\1.80=2P(Z<z)\\P(Z<z)=0.90

    The value of z is ± 1.282.

    The value of x₁ and x₂ are:

    -z=\frac{x_{1}-\mu}{\sigma} \\-1.282=\frac{x_{1}-100}{6}\\x_{1}=100-(6\times1.282)\\=92.308\\\approx92.31       z=\frac{x_{2}-\mu}{\sigma} \\1.282=\frac{x_{2}-100}{6}\\x_{2}=100+(6\times1.282)\\=107.692\\\approx107.70

    Thus, the middle 80% of all heights of 5 year old children fall between 92.31 and 107.70.

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