A count went from 605 to 203. What was the approximate percent decrease? Round the numbers to find an estimate of the percent de
1 answer:
You might be interested in
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Compute the probability that the number of particles withdrawn will be between 235 and 265 as follows: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Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows: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A 10 mL sample is withdrawn.
Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows: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:
Thus, the sample selected must be of size 240.