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 A 5 mL volume of the suspension is withdrawn.

a. What is the probability that the number of particles withdrawn will be between 235 and 265?

b. What is the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52?

c. If a 10 mL sample is withdrawn, what is the probability that the average number per mL of particles in the withdrawn sample is between 48 and 52?

d. How large a sample must be withdrawn so that the average number of particles per mL in the sample it between 48 and 52 with probability 95%?

Step 1 of 8</p>

Let X is the no.of particles in the sample

For one ml the concentration of particles in a suspension is 50

For 5 ml the concentration of particles in a suspension is 5(50)= 250

a) Here we have to find the probability that no. of particles will be between 235 to 265

Here XPoisson(250)

Now X is approximately normal with

mean(=250 and

Standard deviation()=

=15.81

Step 2 of 8</p>

Now we have to find P(235265)

For 235, Z=(

Z=(235 - 250)/15.81

= -0.95

For 265, Z=(

Z=(265 - 250)/15.81

= 0.95

Now we have find P(-0.950.95)=0.82894-0.17106

=0.65788

The probability that no. of particles withdrawn will be between 235 to 265 is 0.65788

Step 3 of 8</p>

b) Here we have to find the probability that the average no. of particles per ml in withdrawn sample will be between 48 and 52

Since the sample is for 5 ml

The average no. of particles will be between 48(5)=240 and 52(5)=260

For 5 ml sample =250

=15.81

Step 4 of 8</p>

Now we have to find P(240260)

For 240, Z=(

Z=(240 - 250)/15.81

= -0.63

For 265, Z=(

Z=(260 - 250)/15.81

= 0.63

Now we have find P(-0.630.63)=0.73565-0.26435

=0.4713

If a 5 sample with drawn the probability that the average no. of particles per ml in the sample will be between 48 and 52 is 0.4713