3 years ago

How do I solve this?

Mathematics

1 answer:

krok68 [10]3 years ago

4 0

3 is 95 and 4 is 85. I believe

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Tcecarenko [31]

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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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3 years ago