Mathematics, 06.03.2020 23:36, fantasticratz2

A researcher is interested in the lengths of Salvelinus fontinalis (brook trout), which are known to be approximately normally distributed with mean 80 centimeters and standard deviation 5 centimeters. To help preserve brook trout populations, some regulatory standards need to be set limiting the size of fish that can be caught. What proportion of fish are larger than 86 centimeters in length

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answered: J3ak06

P(X86)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X86)=P(\frac{X-\mu}{\sigma}\frac{86-\mu}{\sigma})=P(Z\frac{86-80}{5})=P(z1.2)$

And we can find this probability using the complement rule and with the normal standard distribution table:

P(z1.2)=1-P(z

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the lenghts of a population, and for this case we know the distribution for X is given by:

$X \sim N(80,5)$

Where $\mu=80$ and $\sigma=5$

We are interested on this probability

P(X86)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X86)=P(\frac{X-\mu}{\sigma}\frac{86-\mu}{\sigma})=P(Z\frac{86-80}{5})=P(z1.2)$

And we can find this probability using the complement rule and with the normal standard distribution table:

P(z1.2)=1-P(z

answered: Guest

answer: true

step-by-step explanation:

answered: Guest

a. geometricb. g[h] = (3/4)·g[h-1]; g[1] = 384c. g[h] = 384·(3/4)^(h-1)d. g[12] ≈ 16.2183 gallonse. g[12] ≈ 64.8732 quarts

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt