Question

The probability that a person responds to a mailed questionnaire is 0.4. What is the probability that of 25 questionnaires, more than 12 will be returned?

Answer 1

Answer:

Probability = 0.1537

Step-by-step explanation:

Binomial Distribution Condition:

1. There is only 2 outcomes → either return the questionnaire or not return.
2. There is a fixed number of trials → 25 questionnaires.
3. Each trial is independent of all other trials → assuming each person does not persuade others to mail their questionnaire.
4. The probability of success is constant → assuming success rate of 0.4 is constant for each trial.

Although the question's condition meets all requirements for Binomial Distribution, it will take long calculation to find the probability of more than 12 successes. Therefore, we can use Normal Distribution for Binomial Distribution if it also meet these requirements:

1. np > 5
2. nq > 5

where,

• n = no. of trials
• p = success rate
• q = failure rate

Given:

n = 25

p = 0.4

q = 1 - 0.4 = 0.6

x = 12

• np = 25 × 0.4 = 10
• nq = 25 × 0.6 = 15

Since, both np and nq are more than 5, we can use the Normal Distribution for Binomial Distribution → the difference with the standard Normal Distribution is we need to include the Continuity Correction:

• for binomial P(X < x) → normal probability becomes P(V ≤ (x-0.5)
• for binomial P(X ≤ x) → normal probability becomes P(V ≤ (x+0.5)
• for binomial P(X > x) → normal probability becomes P(V ≥ (x+0.5)
• for binomial P(X ≥ x) → normal probability becomes P(V ≥ (x-0.5)

Since the question asks for binomial P(X > 12), then the normal probability become P(V ≥ (12+0.5)) → P(V ≥ 12.5)

Normal Distribution for Binomial Distribution:

$\boxed{mean\ (\mu)=n\cdot p}$

$\boxed{standard\ deviation\ (\sigma)=\sqrt{n\cdot p\cdot q}}$

$\mu=n\cdot p$

  $=25\times0.4$

  $=10$

$\sigma=\sqrt{n\cdot p\cdot q}$

  $=\sqrt{25\times0.4\times0.6}$

  $=\sqrt{6}$

To find the probability of Normal Distribution, first we find the Z-Score:

$\boxed{Z=\frac{x-\mu}{\sigma} }$

$\displaystyle Z=\frac{12.5-10}{\sqrt{6} }$ (using 12.5 instead of 12 → Continuity Correction)

$Z=1.021$

Using the Normal Distribution Table, we can find the probability:

P(V ≥ 12.5) = P(Z ≥ 1.021)

                  = 0.1537

Answer 2

Final answer:

To find the probability of more than 12 out of 25 questionnaires being returned with a return probability of 0.4, one would calculate the cumulative probability of getting 13 or more returns. This typically requires using a binomial probability calculator or a normal approximation because the calculations by hand are intensive for large numbers like 25.

Explanation:

The student's question asks about the probability that more than 12 out of 25 questionnaires are returned when each questionnaire has a probability of 0.4 of being returned. This can be solved using the binomial probability formula, which is P(X = k) = (n choose k) * p^k * (1-p)^(n-k), where 'n' is the number of trials, 'k' is the number of successful outcomes, 'p' is the probability of success on a single trial, and '1-p' is the probability of failure. However, to find the probability of more than 12 returns, we need to calculate the cumulative probability of getting 13 or more returns and then subtract this value from 1, i.e., 1 - P(X ≤ 12). For a large number of trials like 25, this can be quite cumbersome by hand, and typically a normal approximation to the binomial or a binomial probability calculator would be used.

On average, the expected number of respondents (µ) would be n*p, hence 25*0.4 = 10. The standard deviation (σ) of a binomial distribution is √[n*p*(1-p)], which in this case would be √[25*0.4*0.6]. To give an exact numeric answer, one would need to calculate or lookup the cumulative probabilities for each value from 13 to 25.