We read your photo as:

A polling station has a mean arrival rate of 5 voters per 20 minutes. Using the Poisson distribution, calculate the probability of (a) no voters arriving between 3:10 pm and 3:30 pm, and (b) exactly 5

Mathematics

ScanSolve AI Answer

Answers

5 parts
a

0.0067\text{0.0067}

b

0.0019\text{0.0019}

c
5
20
d

0.0819\text{0.0819}

e
21

Step-by-step solution

14 steps

Here are the solutions to the problem:

Part i

Poisson Distribution

The number of voters arriving at a polling station follows a Poisson distribution with a rate of λ=5\lambda = 5 voters per 20 minutes. The probability mass function for a Poisson distribution is P(X=k)=e−λλkk!P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}.

Part a

0.0067\text{0.0067}

No voter will arrive between 3:10 and 3:30 pm

  1. Determine the time interval.

    The interval is 3:30−3:10=203:30 - 3:10 = 20 minutes.

  2. Identify the rate parameter λ\lambda for this interval.

    For a 20-minute interval, λ=5\lambda = 5.

  3. Calculate the probability for k=0k=0.

    P(X=0)=e−5500!=e−5⋅11=e−5P(X=0) = \frac{e^{-5} 5^0}{0!} = \frac{e^{-5} \cdot 1}{1} = e^{-5} P(X=0)≈0.0067379P(X=0) \approx 0.0067379

    The probability that no voter will arrive is 0.0067\boxed{\text{0.0067}}.

Part b

0.0019\text{0.0019}

There will be exactly 5 voters between 9:20-10:20

  1. Determine the time interval.

    The interval is 10:20−9:20=6010:20 - 9:20 = 60 minutes.

  2. Calculate the rate parameter λ′\lambda' for this new interval.

    Since the rate is 5 voters per 20 minutes, for 60 minutes (which is 3×203 \times 20 minutes), the rate is:

    λ′=5×6020=5×3=15\lambda' = 5 \times \frac{60}{20} = 5 \times 3 = 15

  3. Calculate the probability for k=5k=5 with λ′=15\lambda' = 15.

    P(X=5)=e−151555!P(X=5) = \frac{e^{-15} 15^5}{5!} P(X=5)=e−15×759375120P(X=5) = \frac{e^{-15} \times 759375}{120} P(X=5)≈3.05902×10−7×759375120P(X=5) \approx \frac{3.05902 \times 10^{-7} \times 759375}{120} P(X=5)≈0.23226120≈0.0019355P(X=5) \approx \frac{0.23226}{120} \approx 0.0019355

    The probability of exactly 5 voters is 0.0019\boxed{\text{0.0019}}.

Part b(ii)

Geometric Distribution

The probability of getting a wife (success) on any given church service is p=15p = \frac{1}{5}. The number of times he attends church service until he gets a wife follows a Geometric distribution, P(X=k)=(1−p)k−1pP(X=k) = (1-p)^{k-1}p.

Part c5

The mean and variance of the number of times he attends church service

  1. State the probability of success.

    p=15=0.2p = \frac{1}{5} = 0.2.

  2. Calculate the mean E(X)E(X).

    E(X)=1p=10.2=5E(X) = \frac{1}{p} = \frac{1}{0.2} = 5

  3. Calculate the variance Var(X)Var(X).

    Var(X)=1−pp2=1−0.2(0.2)2=0.80.04=20Var(X) = \frac{1-p}{p^2} = \frac{1-0.2}{(0.2)^2} = \frac{0.8}{0.04} = 20

    The mean is 5\boxed{5} and the variance is 20\boxed{20}.

Part d

0.0819\text{0.0819}

The probability that he finds a wife on the 5th church service

  1. Identify the number of trials kk.

    We need the probability that X=5X=5.

  2. Apply the Geometric probability formula.

    P(X=5)=(1−p)5−1p=(1−15)4×15P(X=5) = (1-p)^{5-1}p = \left(1-\frac{1}{5}\right)^4 \times \frac{1}{5} P(X=5)=(45)4×15=256625×15=2563125P(X=5) = \left(\frac{4}{5}\right)^4 \times \frac{1}{5} = \frac{256}{625} \times \frac{1}{5} = \frac{256}{3125} P(X=5)=0.08192P(X=5) = 0.08192

    The probability is 0.0819\boxed{\text{0.0819}}.

Part e21

Find the least integer nn, the number of times he attends church service such that P(X>n)<1%P(X > n) < 1\%.

  1. Set up the inequality for P(X>n)P(X > n).

    For a Geometric distribution, P(X>n)=(1−p)nP(X > n) = (1-p)^n.
    We are given P(X>n)<1%=0.01P(X > n) < 1\% = 0.01.
    So, (1−15)n<0.01\left(1-\frac{1}{5}\right)^n < 0.01.

    (45)n<0.01\left(\frac{4}{5}\right)^n < 0.01 (0.8)n<0.01(0.8)^n < 0.01

  2. Solve for nn using logarithms.

    Take the natural logarithm of both sides:

    nln⁡(0.8)<ln⁡(0.01)n \ln(0.8) < \ln(0.01)

    Since ln⁡(0.8)\ln(0.8) is negative, divide by ln⁡(0.8)\ln(0.8) and reverse the inequality sign:

    n>ln⁡(0.01)ln⁡(0.8)n > \frac{\ln(0.01)}{\ln(0.8)} n>−4.60517−0.22314n > \frac{-4.60517}{-0.22314} n>20.638n > 20.638

  3. Determine the least integer nn.

    The least integer nn that satisfies n>20.638n > 20.638 is 21.
    The least integer nn is 21\boxed{21}.

Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More Mathematics Questions

QA Poisson distribution models voter arrivals at a polling station with an average rate of 5 voters…QThe number of voters arriving at a polling station follows a Poisson distribution with 5 voters in…QFor emergency room receives on average 8 patients per hour. Calculate the mean and variance of…QThe number of emails follows a Poisson distribution with an average rate of 0.2 emails per minute…QA) Count and write the number of votes for each of the towns. b) Draw a histogram showing the…
Browse by topic:Statistics
See all Mathematics questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.