Respuesta :
Answer:
(a) The probability that the randomly selected Rayovac battery will have a test life between 20.4 and 22.2 hours is 0.4178.
(b) The probability that the sample mean of 5 Rayovac battery will have a mean test life between 20.4 and 22.2 hours is 0.6104.
(c) The probability that the sample mean of 15 Rayovac battery will have a mean test life between 20.4 and 22.2 hours is 0.7019.
(d) The probability that the sample mean of 69 Rayovac battery will have a mean test life between 20.4 and 22.2 hours is 0.8708.
Step-by-step explanation:
Let X = the life time of Rayovac batteries.
The lifetime of Rayovac batteries is Normally distributed with mean μ = 20.6 hours and standard deviation σ = 1.47 hours.
The standardized value of the sample mean of X is:
[tex]z=\frac{\bar X-\mu}{\sigma/\sqrt{n}}[/tex]
(a)
Compute the probability that the randomly selected Rayovac battery will have a test life between 20.4 and 22.2 hours as follows:
[tex]P(20.4<X<22.2)=P(\frac{20.4-20.6}{1.47}<\frac{X-\mu}{\sigma}<\frac{22.2-20.6}{1.47})[/tex]
[tex]=P(-0.14<Z<1.09)\\=P(Z<1.09)-P(Z<-0.14)\\=0.8621-0.4443\\=0.4178[/tex]
Thus, the probability that the randomly selected Rayovac battery will have a test life between 20.4 and 22.2 hours is 0.4178.
(b)
Compute the probability that the sample mean of 5 Rayovac battery will have a mean test life between 20.4 and 22.2 hours as follows:
[tex]P(20.4<\bar X<22.2)=P(\frac{20.4-20.6}{1.47/\sqrt{5}}<\frac{\bar X-\mu}{\sigma/\sqrt{n}}<\frac{22.2-20.6}{1.47/\sqrt{5}})[/tex]
[tex]=P(-0.30<Z<2.43)\\=P(Z<2.43)-P(Z<-0.30)\\=0.9925-0.3821\\=0.6104[/tex]
Thus, the probability that the sample mean of 5 Rayovac battery will have a mean test life between 20.4 and 22.2 hours is 0.6104.
(c)
Compute the probability that the sample mean of 15 Rayovac battery will have a mean test life between 20.4 and 22.2 hours as follows:
[tex]P(20.4<\bar X<22.2)=P(\frac{20.4-20.6}{1.47/\sqrt{15}}<\frac{\bar X-\mu}{\sigma/\sqrt{n}}<\frac{22.2-20.6}{1.47/\sqrt{15}})[/tex]
[tex]=P(-0.53<Z<4.22)\\=P(Z<4.22)-P(Z<-0.53)\\=1-0.2981\\=0.7019[/tex]
Thus, the probability that the sample mean of 15 Rayovac battery will have a mean test life between 20.4 and 22.2 hours is 0.7019.
(d)
Compute the probability that the sample mean of 69 Rayovac battery will have a mean test life between 20.4 and 22.2 hours as follows:
[tex]P(20.4<\bar X<22.2)=P(\frac{20.4-20.6}{1.47/\sqrt{69}}<\frac{\bar X-\mu}{\sigma/\sqrt{n}}<\frac{22.2-20.6}{1.47/\sqrt{69}})[/tex]
[tex]=P(-1.13<Z<9.04)\\=P(Z<9.04)-P(Z<-1.13)\\=1-0.1292\\=0.8708[/tex]
Thus, the probability that the sample mean of 69 Rayovac battery will have a mean test life between 20.4 and 22.2 hours is 0.8708.