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Vinil7 [7]
3 years ago
9

In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal dist

ribution to estimate the requested probabilities. Ocean fishing for billfish is very popular in the Cozumel region of Mexico. In the Cozumel 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 23 strikes. Find the following probabilities. (Round your answers to four decimal places.) (a) 12 or fewer fish were caught (b) 5 or more fish were caught (c) between 5 and 12 fish were caught
Mathematics
1 answer:
Tcecarenko [31]3 years ago
4 0

Answer:

a. 0.7291

b. 0.9968

c. 0.7259

Step-by-step explanation:

a. np and n(1-p) can be calculated as:

np=23\times 0.48\\\\=11.04\\\\n(1-p)=23(1-0.52)\\\\=11.96

#Both np and np(1-p) are greater than 5, hence, normal approximation is most appropriate:

\mu_x=11.04\\\\\sigma^2=np(1-p)=0.48\times 0.52\times 23=5.7408

#Define Y:

Y~(11.04,5.7408)

P(X\leq 12)\approx P(Y\leq 12.5)\\\\P(Z\leq \frac{(12.5-11.04)}{\sqrt{5.7408}})=\\\\=1-0.2709\\\\=0.7291

Hence, the probability of 12 or fewer is 0.8291

b. The  probability that 5 or more fish were caught.

#Using normal approximation:

P(X \geq 5) \approx P(Y \geq 4.5) = P(Z \geq\frac{ (4.5-11.04)}{\sqrt{(5.7408)}})\\\\=1-0.0032\\\\=0.9968

Hence, the probability of catching 5+ is 0.9968

c. The probability of between 5 and 12 is calculated as;

-From b above P(X\geq 5)=0.9968 and a ,P(X\leq 12)=0.7291

P(5\leq X\leq 120\approx P(4.5\leq Y\leq  12.5)\\\\=0.7291-(1-0.9968)\\\\=0.7259

Hence, the probability of between 5 and 12 is 0.7259

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