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3 years ago

8

As part of a sixth-grade class project, the teacher brings to class a large jar containing 200 gumballs of two different colo

rs: red and green. Andy is asked to draw a sample of his own choosing and estimate the number of red gumballs in the jar. Andy draws a sample of 4040 gumballs, of which 88 are red and 3232 are green. Use Andy's sample to estimate the number of red gumballs in the jar.

1 answer:

Rudiy273 years ago

7 0

Answer:

There would be 40 red gumballs

Step-by-step explanation:

Since, out of 40 gumballs there are 8 red gumballs,

So, the ratio of red gumballs and total gumballs = $\frac{8}{40}$

$=\frac{1}{5}$

Now, let there are x red gumballs in 200 gumballs,

The ratio of red gumballs and total gumballs = $\frac{x}{200}$

$\implies \frac{x}{200}=\frac{1}{5}$

$x =\frac{200}{5}=40$

Hence, the possible number of red gumballs in the jar would be 40.

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The time between arrivals of customers at the drive-up window of a bank follows an exponential probability distribution with a m

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Answer:

a) 50.34% probability that the arrival time between customers will be 7 minutes or less.

b) 24.42% probability that the arrival time between customers will be between 3 and 7 minutes

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

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The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

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The probability of finding a value higher than x is:

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Mean of 10 minutes:

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A. What is the probability that the arrival time between customers will be 7 minutes or less?

$P(X \leq x) = 1 - e^{-\mu x}$

$P(X \leq 7) = 1 - e^{-0.1*7} = 0.5034$

50.34% probability that the arrival time between customers will be 7 minutes or less.

B. What is the probability that the arrival time between customers will be between 3 and 7 minutes?

$P(3 \leq X \leq 7) = P(X \leq 7) - P(X \leq 3)$

$P(X \leq x) = 1 - e^{-\mu x}$

$P(X \leq 7) = 1 - e^{-0.1*7} = 0.5034$

$P(X \leq 3) = 1 - e^{-0.1*3} = 0.2592$

$P(3 \leq X \leq 7) = P(X \leq 7) - P(X \leq 3) = 0.5034 - 0.2592 = 0.2442$

24.42% probability that the arrival time between customers will be between 3 and 7 minutes

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What is the range of the equation

a_sh-v [17]

The range of the equation is $y>2$

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