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Ainat [17]
3 years ago
10

A normal random variable with mean and standard deviationboth equal to 10 degrees Celsius. What is the probability that the temp

erature at a randomly chosen time will be less than or equal to 59 degrees Fahrenheit?
Mathematics
1 answer:
Korvikt [17]3 years ago
4 0

Answer:

0.57142

Step-by-step explanation:

A normal random variable with mean and standard deviation both equal to 10 degrees Celsius. What is the probability that the temperature at a randomly chosen time will be less than or equal to 59 degrees Fahrenheit?

We are told that the Mean and Standard deviation = 10°C

We convert to Fahrenheit

(10°C × 9/5) + 32 = 50°F

Hence, we solve using z score formula

z = (x-μ)/σ, where

x is the raw score = 59 °F

μ is the population mean = 50 °F

σ is the population standard deviation = 50 °F

z = 59 - 50/50

z = 0.18

Probability value from Z-Table:

P(x ≤59) = 0.57142

The probability that the temperature at a randomly chosen time will be less than or equal to 59 degrees Fahrenheit

is 0.57142

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Mazyrski [523]

Answer:

The population in 40 years will be 1220.

Step-by-step explanation:

The population of a town grows at a rate proportional to the population present at time t.

This means that:

P(t) = P(0)e^{rt}

In which P(t) is the population after t years, P(0) is the initial population and r is the growth rate.

The initial population of 500 increases by 25% in 10 years.

This means that P(0) = 500, P(10) = 1.25*500 = 625

We apply this to the equation and find t.

P(t) = P(0)e^{rt}

625 = 500e^{10r}

e^{10r} = \frac{625}{500}

e^{10r} = 1.25

Applying ln to both sides

\ln{e^{10r}} = \ln{1.25}

10r = \ln{1.25}

r = \frac{\ln{1.25}}{10}

r = 0.0223

So

P(t) = 500e^{0.0223t}

What will be the population in 40 years

This is P(40).

P(t) = 500e^{0.0223t}

P(40) = 500e^{0.0223*40} = 1220

The population in 40 years will be 1220.

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