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Triss [41]
3 years ago
7

A researcher is interested in the lengths of 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 for limiting the size of fish that can be caught. What is the probability of catching a brook trout less than 72 centimeters in length
Mathematics
1 answer:
aliya0001 [1]3 years ago
6 0

Answer:

P(X

And we can find this probability with the normal standard distirbution or excel and we got:

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(X

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(X

And we can find this probability with the normal standard distirbution or excel and we got:

P(z

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8 0
3 years ago
Elizabeth lives in San Francisco and works in Mountain View. In the morning, she has 3 transportation options (take a bus, a cab
mina [271]

Answer:

If Elizabeth randomly chooses her ride in the morning and in the evening, 2/3 is the probability that she'll use a cab exactly one time

3 0
2 years ago
Read 2 more answers
Samples of emissions from three suppliers are classified for conformance to air-quality specifications. The results from 100 sam
tigry1 [53]

Answer:

P(A) = \frac{30}{100}

P(B) = \frac{77}{100}

P(A\ n\ B) = \frac{22}{100}

P(A\ u\ B) = \frac{85}{100}

Step-by-step explanation:

Given

See attachment for proper format of table

n = 100 --- Sample

A = Supplier 1

B = Conforms to specification

Solving (a): P(A)

Here, we only consider data in sample 1 row.

In this row:

Yes = 22 and No = 8

So, we have:

n(A) = Yes + No

n(A) = 22 + 8

n(A) = 30

P(A) is then calculated as:

P(A) = \frac{n(A)}{Sample}

P(A) = \frac{30}{100}

Solving (b): P(B)

Here, we only consider data in the Yes column.

In this column:

(1) = 22    (2) = 25 and (3) = 30

So, we have:

n(B) = (1) + (2) + (3)

n(B) = 22 + 25 + 30

n(B) = 77

P(B) is then calculated as:

P(B) = \frac{n(B)}{Sample}

P(B) = \frac{77}{100}

Solving (c): P(A n B)

Here, we only consider the similar cell in the yes column and sample 1 row.

This cell is: [Supplier 1][Yes]

And it is represented with; n(A n B)

So, we have:

n(A\ n\ B) = 22

The probability is then calculated as:

P(A\ n\ B) = \frac{n(A\ n\ B)}{Sample}

P(A\ n\ B) = \frac{22}{100}

Solving (d): P(A u B)

This is calculated as:

P(A\ u\ B) = P(A) + P(B) - P(A\ n\ B)

This gives:

P(A\ u\ B) = \frac{30}{100} + \frac{77}{100} - \frac{22}{100}

Take LCM

P(A\ u\ B) = \frac{30+77-22}{100}

P(A\ u\ B) = \frac{85}{100}

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3 years ago
Please Help!<br><br> Car X travels 174 miles in 3 hours.
DiKsa [7]

Answer:

The graph =

<h3>y = 58x</h3>

Step-by-step explanation:

(a)

distance covered (y) = 174 miles

time taken (x) = 3 hours

Therefore the points = (3, 174) and (0, 0) = stationary position

Find the gradient (m)

m = 174-0 / 3-0

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Therefore the equation =

<h3>y = 58x</h3>

(distance = 58 × time)

For the distance = 174 and time = 3 hours, the equation to find either distance or time = y = 58x

(b) the graph is constant, because the time and distance are also contant when the speed used is the same.

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iogann1982 [59]
C. 728

14x40=560
14+7=21
21x8=168
560+168=728
3 0
2 years ago
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