Respuesta :
Answer:
(a) 0.6579
(b) 0.2961
(c) 0.3108
(d) 240
Step-by-step explanation:
The random variable X can be defined as the number of particles in a suspension.
The concentration of particles in a suspension is 50 per ml.
Then in 5 mL volume of the suspension the number of particles will be,
5 × 50 = 250.
The random variable X thus follows a Poisson distribution with parameter, λ = 250.
The Poisson distribution with parameter λ, can be approximated by the Normal distribution, when λ is large say λ > 10.
The mean of the approximated distribution of X is:
μ = λ = 250
The standard deviation of the approximated distribution of X is:
σ = √λ = √250 = 15.8114
Thus, [tex]X\sim N(250, 250)[/tex]
(a)
Compute the probability that the number of particles withdrawn will be between 235 and 265 as follows:
[tex]P(235<X<265)=P(\frac{235-250}{15.8114}<\frac{X-\mu}{\sigma}<\frac{265-250}{15.8114})[/tex]
[tex]=P(-0.95<Z<0.95)\\=P(Z<0.95)-P(Z<-0.95)\\=P(Z<0.95)-[1-P(Z<0.95)]\\=2P(Z<0.95)-1\\=(2\times 0.82894)-1\\=0.65788\\\approx 0.6579[/tex]
Thus, the value of P (235 < X < 265) = 0.6579.
(b)
Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows:
[tex]P(48<\bar X<52)=P(\frac{48-50}{15.8114/\sqrt{5}}<\frac{\bar X-\mu}{\sigma/\sqrt{n}}<\frac{52-50}{15.8114/\sqrt{5}})[/tex]
[tex]=P(-0.28<Z<0.28)\\=P(Z<0.28)-P(Z<-0.28)\\=P(Z<0.28)-[1-P(Z<0.28)]\\=2P(Z<0.28)-1\\=(2\times 0.64803)-1\\=0.29606\\\approx 0.2961[/tex]
Thus, the value of [tex]P(48<\bar X<52)=0.2961[/tex].
(c)
A 10 mL sample is withdrawn.
Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows:
[tex]P(48<\bar X<52)=P(\frac{48-50}{15.8114/\sqrt{10}}<\frac{\bar X-\mu}{\sigma/\sqrt{n}}<\frac{52-50}{15.8114/\sqrt{10}})[/tex]
[tex]=P(-0.40<Z<0.40)\\=P(Z<0.40)-P(Z<-0.40)\\=P(Z<0.40)-[1-P(Z<0.40)]\\=2P(Z<0.40)-1\\=(2\times 0.65542)-1\\=0.31084\\\approx 0.3108[/tex]
Thus, the value of [tex]P(48<\bar X<52)=0.3108[/tex].
(d)
Let the sample size be n.
[tex]P(48<\bar X<52)=P(\frac{48-50}{15.8114/\sqrt{n}}<\frac{\bar X-\mu}{\sigma/\sqrt{n}}<\frac{52-50}{15.8114/\sqrt{n}})[/tex]
[tex]0.95=P(-z<Z<z)\\0.95=P(Z<z)-P(Z<-z)\\0.95=P(Z<z)-[1-P(Z<z)]\\0.95=2P(Z<z)-1\\P(Z<z)=\frac{1.95}{2}\\\\P(Z<z)=0.975[/tex]
The value of z for this probability is,
z = 1.96
Compute the value of n as follows:
[tex]z=\frac{\bar X-\mu}{\sigma/\sqrt{n}}\\\\1.96=\frac{48-50}{15.8114/\sqrt{n}}\\\\n=[\frac{1.96\times 15.8114}{48-50}]^{2}\\\\n=240.1004\\\\n\approx 241[/tex]
Thus, the sample selected must be of size 240.
