Question

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 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

Answer 1

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