SOLUTION: 5. The number of white corpuscles on a slide has a Poisson distribution with mean 3.6. a. Find the most likely number of white corpuscles on a slide. b. Calculate correct to thre

Question 1192083: 5. The number of white corpuscles on a slide has a Poisson distribution with mean 3.6.
a. Find the most likely number of white corpuscles on a slide.
b. Calculate correct to three decima places the probability of obtaining this number.
c. If such two slides are prepared, what is the probability, correct to three decimal places, of obtaining at least two white corpuscles in total on the two slides?
6. At Foodland Supermarket 65% of customers pay by debit card. Find the probability that in a randomly selected sample of twenty customers.
a. Exactly five pay by debit card
b. More than eighteen pay by cash.
(give answers to 3 decimal places)

Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Here's the solution:
**Problem 1: White Corpuscles**
**a. Most Likely Number:**
For a Poisson distribution, the most likely number is the integer part of the mean (λ). In this case, λ = 3.6. Therefore, the most likely number of white corpuscles is 3.
**b. Probability of Obtaining the Most Likely Number:**
Use the Poisson probability formula:
P(x) = (e^(-λ) * λ^x) / x!
P(3) = (e^(-3.6) * 3.6^3) / 3!
P(3) ≈ (0.0273 * 46.656) / 6
P(3) ≈ 0.217
The probability of obtaining 3 white corpuscles is approximately 0.217.
**c. Probability of at Least Two White Corpuscles in Total on Two Slides:**
* The mean for two slides is 2 * 3.6 = 7.2.
* We want P(X ≥ 2). It's easier to calculate the complement: P(X ≥ 2) = 1 - P(X < 2) = 1 - [P(0) + P(1)]
* P(0) = (e^(-7.2) * 7.2^0) / 0! ≈ 0.0007
* P(1) = (e^(-7.2) * 7.2^1) / 1! ≈ 0.0051
* P(X ≥ 2) = 1 - (0.0007 + 0.0051)
* P(X ≥ 2) = 1 - 0.0058
* P(X ≥ 2) ≈ 0.994
The probability of at least two white corpuscles in total is approximately 0.994.
**Problem 2: Foodland Supermarket**
This is a binomial probability problem.
* n = 20 (number of customers)
* p = 0.65 (probability of paying by debit card)
* q = 1 - p = 0.35 (probability of paying by cash)
**a. Exactly Five Pay by Debit Card:**
P(X = 5) = (20C5) * p^5 * q^15
P(X = 5) = (20! / (5! * 15!)) * (0.65)^5 * (0.35)^15
P(X = 5) ≈ 15504 * 0.1160 * 1.406 x 10⁻⁸
P(X = 5) ≈ 0.002
The probability of exactly five customers paying by debit card is approximately 0.002.
**b. More Than Eighteen Pay by Cash:**
P(X > 18) = P(X = 19) + P(X = 20)
P(X = 19) = (20C19) * (0.65)^19 * (0.35)^1 ≈ 20 * 0.00017 * 0.35 ≈ 0.001
P(X = 20) = (20C20) * (0.65)^20 * (0.35)^0 ≈ 1 * 0.00037 * 1 ≈ 0.000
P(X > 18) ≈ 0.001 + 0.000 = 0.001
The probability of more than eighteen customers paying by cash is approximately 0.001.

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