Answer:
(a) 0.6579
(b) 0.2961
(c) 0.3108
(d) 240
Step-by-step explanation:
The random variable <em>X</em> 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 <em>X</em> thus follows a Poisson distribution with parameter, <em>λ</em> = 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, 
(a)
Compute the probability that the number of particles withdrawn will be between 235 and 265 as follows:
Thus, the value of P (235 < <em>X</em> < 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Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows:
Thus, the value of 
.
(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A 10 mL sample is withdrawn.
Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows:
Thus, the value of 
.
(d)
Let the sample size be <em>n</em>.
The value of <em>z</em> for this probability is,
<em>z</em> = 1.96
Compute the value of <em>n</em> as follows:
![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](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B%5Cbar%20X-%5Cmu%7D%7B%5Csigma%2F%5Csqrt%7Bn%7D%7D%5C%5C%5C%5C1.96%3D%5Cfrac%7B48-50%7D%7B15.8114%2F%5Csqrt%7Bn%7D%7D%5C%5C%5C%5Cn%3D%5B%5Cfrac%7B1.96%5Ctimes%2015.8114%7D%7B48-50%7D%5D%5E%7B2%7D%5C%5C%5C%5Cn%3D240.1004%5C%5C%5C%5Cn%5Capprox%20241)
Thus, the sample selected must be of size 240.
