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Lera25 [3.4K]
2 years ago
9

The time that a butterfly lives after emerging from its chrysalis can be modelled by a random variable tt, the model here taking

the probability that a butterfly survives for more than tt days as
Mathematics
1 answer:
4vir4ik [10]2 years ago
3 0
<span>Answer: The probabilities that either a) the butterfly dies within 7 days or b) the butterfly lives longer than 7 days must add up to one. We know the second probability (lives longer than 7 days)--it is given. So now set up an equation and solve: P(t > 7) = 36 / 13² = 36/169 --> 36/169 + x = 1 --> x = 1 - 36/169 = 133/169 ~ 0.7870 or 78.70% b) 7% is the probability (the expected value) that a butterfly will survive more than t days. Again, this formula is given: 36/(t + 6)² = 0.07 --> (t + 6)² = 36/.07 --> t = âš(36/.07) - 6 ~ 16.68 days This is the value of t that will give a probability of 7%. c) This one is a little trickier. Technically the mean lifetime should be calculated as: â«t * p(t)dt where p(t) is the probability distribution function But we have a cumulative probability distribution. I think we need to find the probability distribution from this cumulative one. Which is easy because the cumulative distribution function is an integral of the probability distribution function: P(t) = â«p(x)dx, from x = t, to x = âž Now hopefully it's easy to see that P(t) is an anti-derivative of p(x). Said the other way p(x) is the derivative of P(t): P'(t) = -72/(t + 6)Âł --> but p(t) shouldn't be negative--also you can convince yourself that it should be positive by using the above value (from t --> âž...you get -0 - a negative = a positive) p(t) = 72/(t + 6)Âł Now we can calculate the mean lifetime by integrating: 72â«t/(t + 6)Âłdt --> there's a very easy u-substitution to use here: u = t + 6 --> du = dt --> t = u - 6 --> â«(u - 6)/uÂłdu = â«(1/u² - 6/uÂł)du = -1/u + 3/u² + C --> now evaluate from t = 0 to t = âž or from u = 0 + 6 = 6 to u = âž 0 - (-1/6 + 3/36) = 1/6 - 1/12 = 1/12 --> multiply by the 72 72/12 = 6 days</span>
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The time between failures for an electrical appliance is exponentially distributed with a mean of 25 months. What is the probabi
katrin [286]

Answer:

The probability that the next failure will not occur before 30 months have elapsed is 0.0454

Step-by-step explanation:

Using Poisson distribution  where

t= number of units of time

x= number of occurrences in t units of time

λ= average number of occurrences per unit of time

P(x;λt) = e raise to power (-λt)  multiplied by λtˣ divided by x!

here λt = 25

x= 30

P(x= 30) = 25³⁰e⁻²⁵/ 30!

P (x= 30) = 8.67 E41 * 1.3887 E-11/30!    (where E= exponent)

P (x=30) = 1.204 E31/30!

Solving it with a statistical calculator would give

P (x=30) = 0.0454

The probability that the next failure will not occur before 30 months have elapsed is 0.0454

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If AD=20 and AC=3x+4, find the value of x. Then find AC and DC.<br><br> I don't understand.
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The size of fish is very important to commercial fishing. A study conducted in 2012 found the length of Atlantic cod caught in n
devlian [24]

Answer:

The length of the longest 15% of Atlantic cod in this area is 53.79cm, roundeed to 2 decimal places.

Step-by-step explanation:

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

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

Z = \frac{X - \mu}{\sigma}

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.

In this problem, we have that:

A study conducted in 2012 found the length of Atlantic cod caught in nets in Karlskrona to have a mean of 49.9 cm and a standard deviation of 3.74 cm, so \mu = 49.9, \sigma = 3.74.

What is the length in cm of the longest 15% of Atlantic cod in this area?

We have to find the value of X for the value of Z that has a pvalue of 0.85.

Looking at the zscore table, we have that Z = 1.04 has a pvalue of 0.8508. So, we have to find the value of X when Z = 1.04, \mu = 49.9, \sigma = 3.74.

So

Z = \frac{X - \mu}{\sigma}

1.04 = \frac{X - 49.9}{3.74}

X - 49.9 = 3.8896

X = 53.7896

The length of the longest 15% of Atlantic cod in this area is 53.79cm, roundeed to 2 decimal places.

8 0
2 years ago
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