Respuesta :
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:
[tex]P (X>89.2)=P(\frac{X-\mu}{\sigma}>\frac{89.2-100}{6})\\=P(Z>-1.80)\\=P(Z<1.8)\\=0.9641[/tex]
Thus, the value of P (X > 89.2) is 0.9641.
(b)
Compute the value of P (X < 109.78) as follows:
[tex]P (X<109.78)=P(\frac{X-\mu}{\sigma}<\frac{109.78-100}{6})\\=P(Z<1.63)\\=0.9484[/tex]
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)
[tex]=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[/tex]
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)
[tex]=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[/tex]
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:
[tex]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[/tex]
Thus, the value of P (X > 103) is 0.3085.
(f)
Compute the value of P (X < 98.2) as follows:
[tex]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[/tex]
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)
[tex]=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[/tex]
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:
[tex]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[/tex]
The value of z is ± 1.282.
The value of x₁ and x₂ are:
[tex]-z=\frac{x_{1}-\mu}{\sigma} \\-1.282=\frac{x_{1}-100}{6}\\x_{1}=100-(6\times1.282)\\=92.308\\\approx92.31[/tex] [tex]z=\frac{x_{2}-\mu}{\sigma} \\1.282=\frac{x_{2}-100}{6}\\x_{2}=100+(6\times1.282)\\=107.692\\\approx107.70[/tex]
Thus, the middle 80% of all heights of 5 year old children fall between 92.31 and 107.70.
