Respuesta :
Answer:
(a) 0.7967
(b) 0.6826
(c) 0.3707
(d) 0.9525
(e) 0.1587
Step-by-step explanation:
The random variable X follows a Normal distribution with mean μ = 10 and variance σ² = 36.
(a)
Compute the value of P (X > 5) as follows:
[tex]P(X>5)=P(\frac{x-\mu}{\sigma}>\frac{5-10}{\sqrt{36}})\\=P(Z>-0.833)\\=P(Z<0.83)\\=0.7967[/tex]
Thus, the value of P (X > 5) is 0.7967.
(b)
Compute the value of P (4 < X < 16) as follows:
[tex]P(4<X<16)=P(\frac{4-10}{\sqrt{36}}<\frac{x-\mu}{\sigma}<\frac{16-10}{\sqrt{36}})\\=P(-1<Z<1)\\=P(Z<1)-P(Z<-1)\\=P(Z<1)-1+P(Z<1)\\=2P(Z<1)-1\\=(2\times0.8413)-1\\=0.6826[/tex]
Thus, the value of P (4 < X < 16) is 0.6826.
(c)
Compute the value of P (X < 8) as follows:
[tex]P(X<8)=P(\frac{x-\mu}{\sigma}<\frac{8-10}{\sqrt{36}})\\=P(Z<-0.33)\\=1-P(Z<0.33)\\=1-0.6293\\=0.3707[/tex]
Thus, the value of P (X < 8) is 0.3707.
(d)
Compute the value of P (X < 20) as follows:
[tex]P(X<20)=P(\frac{x-\mu}{\sigma}<\frac{20-10}{\sqrt{36}})\\=P(Z<1.67)\\=0.9525[/tex]
Thus, the value of P (X < 20) is 0.9525.
(e)
Compute the value of P (X > 16) as follows:
[tex]P(X>16)=P(\frac{x-\mu}{\sigma}>\frac{16-10}{\sqrt{36}})\\=P(Z>1)\\=1-P(Z<1)\\=1-0.8413\\=0.1587[/tex]
Thus, the value of P (X > 16) is 0.1587.
**Use a z-table for the probabilities.
