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skad [1K]
3 years ago
9

A bacteria culture starts with 200 bacteria and grows at a rate proportional to its size. After 6 hours there will be 1200 bacte

ria (1) Express the population after I hours as a function of t. population: p(tepe (1.066-21) (unction of t) (b) What will be the population after 7 hours? 348125.2 (c) How long will it take for the population to reach 1750 ? Note: You can earn partial credit on this problem.
Mathematics
1 answer:
Harman [31]3 years ago
6 0

Answer:

(1) Let, P represents the size of the bacteria culture in t hours,

According to the question,

\frac{dP}{dt}\propto P

\frac{dP}{dt}=kP

\frac{dP}{P}=kdt

By integrating,

\ln P=kt + C

P=e^{kt+C}

P=e^{kt}.e^C

P=P_0 e^{kt} Where e^C=P_0,

We have,

at t = 0, P = 200,

\implies 200=P_0 e^0\implies P_0 = 200

at t = 6, P = 1200

1200=200 e^{6k}\implies k = 0.299

Hence, the required function would be,

P=200 e^{0.299t}

(2) if t = 7,

The population would be,

P=200 e^{0.299\times 7}=1621.84126567\approx 1622

(3) If P = 1750,

1750=200 e^{0.299t}\implies t = 7.254

Hence, it will take about 8 years for the population to reach 1750.

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Step-by-step explanation:

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

Normal probability distribution

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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.

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

In this problem, we have that:

\mu = 26, \sigma = 6, n = 62, s = \frac{6}{\sqrt{62}} = 0.762

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What is the likelihood the sample mean is at least $27.00?

This is 1 subtracted by the pvalue of Z when X = 27. So

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

By the Central Limit Theorem

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

Z = \frac{27 - 26}{0.762}

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Z = 1.31 has a pvalue of 0.9049

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(b)

What is the likelihood the sample mean is greater than $25.00 but less than $27.00?

This is the pvalue of Z when X = 27 subtracted by the pvalue of Z when X = 25. So

X = 27

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

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

Z = \frac{25 - 26}{0.762}

Z = -1.31

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Between the 5th and the 95th percentile.

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X when Z has a pvalue of 0.05. So X when Z = -1.645

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X - 26 = -1.645*0.762

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X when Z has a pvalue of 0.95. So X when Z = 1.645

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1.645 = \frac{X - 26}{0.762}

X - 26 = 1.645*0.762

X = 27.25

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