6. Suppose Steve goes fishing near the nuclear power plant at Hawkins. He’s interested in catching King Salmons and Walleyes. Assume the following: • All species of fish in the lake have weights that are normally distributed. • The weight of the King Salmons are i.i.d. ∼ Normal with µK = 150 lbs and σK = 10 lbs. Let K be the weight of a randomly caught King Salmon. • The weight of the Walleyes are i.i.d. ∼ Normal with µW = 51 lbs and σW = 9 lbs. Let W be the weight of a randomly caught Walleye. (a) (3 points) Suppose Steve catches 4 King Salmons at random. What is the probability that the total weight of the King Salmons caught is greater than 575 lbs?

Respuesta :

Answer:

89.44% probability that the total weight of the King Salmons caught is greater than 575 lbs

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For sums of size n from a population, the mean is [tex]\mu*n[/tex] and the standard deviation is [tex]\sigma\sqrt{n}[/tex]

The weight of the King Salmons are i.i.d. ∼ Normal with µK = 150 lbs and σK = 10 lbs. 4 king salmons.

So [tex]\mu = 4*150 = 600, \sigma = 10\sqrt{4} = 20[/tex]

What is the probability that the total weight of the King Salmons caught is greater than 575 lbs?

This is 1 subtracted by the pvalue of Z when X = 575. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{575 - 600}{20}[/tex]

[tex]Z = -1.25[/tex]

[tex]Z = -1.25[/tex] has a pvalue of 0.1056

1 - 0.1056 = 0.8944

89.44% probability that the total weight of the King Salmons caught is greater than 575 lbs