Respuesta :
Answer:
(a) 0.4945.
(b) 0.3643.
(c) 0.5965.
(d) 0.0869.
(e) 0.9974.
Step-by-step explanation:
Let X = readings on thermometers in a room.
The random variable X follows a Normal distribution with mean, μ = 0°C and standard deviation, σ = 1.00°C.
(a)
Compute the probability that the reading on the thermometer is between 0°C an 2.54°C as follows:
[tex]P(0<X<2.54)=P(\frac{0-0}{1}<\frac{X-\mu}{\sigma}<\frac{2.54-0}{1})\\=P(0<Z<2.54)\\=P(Z<2.54)-P(Z<0)\\=0.9945-0.50\\=0.4945[/tex]
*Use a z-table.
Thus, the probability that the reading on the thermometer is between 0°C an 2.54°C is 0.4945.
(b)
Compute the probability that the reading on the thermometer is between -1.10°C an 0°C as follows:
[tex]P(-1.1<X<0)=P(\frac{-1.1-0}{1}<\frac{X-\mu}{\sigma}<\frac{0-0}{1})\\=P(-1.1<Z<0)\\=P(Z<0)-P(Z<-1.1)\\=0.50-0.1357\\=0.3643[/tex]
*Use a z-table.
Thus, the probability that the reading on the thermometer is between -1.10°C an 0°C is 0.3643.
(c)
Compute the probability that the reading on the thermometer is between -0.38°C an 1.63°C as follows:
[tex]P(-0.38<X<1.63)=P(\frac{-0.38-0}{1}<\frac{X-\mu}{\sigma}<\frac{1.63-0}{1})\\=P(-0.38<Z<1.63)\\=P(Z<1.63)-P(Z<-0.38)\\=0.9485-0.3520\\=0.5965[/tex]
*Use a z-table.
Thus, the probability that the reading on the thermometer is between -0.38°C an 1.63°C is 0.5965.
(d)
Compute the probability that the reading on the thermometer is less than -1.36°C as follows:
[tex]P(X<-1.36)=P(\frac{A-\mu}{\sigma}<\frac{-1.36-0}{1})=P(Z<-1.36)=0.0869[/tex]
Thus, the probability that the reading on the thermometer is less than -1.36°C is 0.0869.
(e)
Compute the probability that the reading on the thermometer is greater than -2.79°C as follows:
[tex]P(X>-2.79)=P(\frac{A-\mu}{\sigma}>\frac{-2.79-0}{1})=P(Z<-2.79)=0.9974[/tex]
Thus, the probability that the reading on the thermometer is greater than -2.79°C is 0.9974.
