The concentration of particles in a suspension is 50 per mL. A 5 mL volume of the suspension is withdrawn. a. What is the probability that the number of particles withdrawn will be between 235 and 265? b. What is the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52? c. If a 10 mL sample is withdrawn, what is the probability that the average number per mL of particles in the withdrawn sample is between 48 and 52? d. How large a sample must be withdrawn so that the average number of particles per mL in the sample is between 48 and 52 with probability 95%?

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Compute the probability that the number of particles withdrawn will be between 235 and 265 as follows: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Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows: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A 10 mL sample is withdrawn.

Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows: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.

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