Math Problem Statement

In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to estimate the requested probabilities.

Ocean fishing for billfish is very popular in a certain region of Mexico. In this region about 48% of strikes (while trolling) resulted in a catch. Suppose that on a given day a fleet of fishing boats got a total of 21 strikes. Find the following probabilities. (Round your answers to four decimal places.)

12 or fewer fish were caught

Solution

To determine whether it's appropriate to use the normal approximation to the binomial distribution, we need to check the following two conditions:

  1. np5np \geq 5 (where nn is the number of trials and pp is the probability of success)
  2. n(1p)5n(1-p) \geq 5 (where 1p1-p is the probability of failure)

Given:

  • n=21n = 21 (number of trials, or the total number of strikes)
  • p=0.48p = 0.48 (probability of success, or the probability of catching a fish)

Now, let's check the conditions:

  1. np=21×0.48=10.085np = 21 \times 0.48 = 10.08 \geq 5
  2. n(1p)=21×(10.48)=21×0.52=10.925n(1-p) = 21 \times (1 - 0.48) = 21 \times 0.52 = 10.92 \geq 5

Both conditions are satisfied, so it's appropriate to use the normal approximation.

Step 1: Define the problem in terms of the binomial distribution

  • The binomial distribution has a mean μ=np=21×0.48=10.08\mu = np = 21 \times 0.48 = 10.08
  • The standard deviation σ=np(1p)=21×0.48×0.52=5.24162.2882\sigma = \sqrt{np(1-p)} = \sqrt{21 \times 0.48 \times 0.52} = \sqrt{5.2416} \approx 2.2882

Step 2: Apply the continuity correction

To use the normal distribution as an approximation to the binomial distribution, we need to apply a continuity correction when finding the probability of 12 or fewer fish. We approximate P(X12)P(X \leq 12) by calculating P(X12.5)P(X \leq 12.5).

Step 3: Standardize the variable

We now convert X=12.5X = 12.5 to a z-score:

z=Xμσ=12.510.082.2882=2.422.28821.0576z = \frac{X - \mu}{\sigma} = \frac{12.5 - 10.08}{2.2882} = \frac{2.42}{2.2882} \approx 1.0576

Step 4: Use the standard normal distribution

Using a standard normal distribution table or calculator, we find the probability corresponding to z=1.0576z = 1.0576.

P(Z1.0576)0.8553P(Z \leq 1.0576) \approx 0.8553

Final Answer:

The probability that 12 or fewer fish were caught is approximately 0.85530.8553.


Would you like more details or clarifications?


Here are five related questions to explore this concept further:

  1. How does the continuity correction affect the approximation in a binomial distribution?
  2. What would happen if the sample size nn were smaller in this problem?
  3. How is the normal distribution derived from the binomial distribution?
  4. What are other common applications of the normal approximation to the binomial distribution?
  5. How would the probability change if the success rate (probability of catching a fish) were higher?

Tip: Always check both conditions np5np \geq 5 and n(1p)5n(1-p) \geq 5 before using the normal approximation to the binomial distribution.

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Math Problem Analysis

Mathematical Concepts

Binomial Distribution
Normal Approximation
Probability

Formulas

np (mean of binomial distribution)
sqrt(np(1-p)) (standard deviation of binomial distribution)
z = (X - μ) / σ (z-score formula)
Continuity Correction

Theorems

Central Limit Theorem
Normal Approximation to the Binomial Distribution

Suitable Grade Level

Grades 11-12