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4 years ago

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A boy has color blindness and has trouble distinguishing blue and green. There are 75 blue pens and 25 green pens mixed together

in a box. Given that he picks up a blue pen, there is a 80% chance that he thinks it is a blue pen and a 20% chance that he thinks it is a green pen. Given that he picks a green pen, there is an 90% chance that he thinks it is a green pen and a 10% chance that he thinks it is a blue pen. Assume that the boy randomly selects one of the pens from the box.
a) (3pts) What is the probability that he picks up a blue pen and recognizes it as blue?

b) (3 pts) What is the probability that he chooses a pen and thinks it is blue?

c) (4pts) Given that he thinks he chose a blue pen, what is the probability that he actually chose a blue pen?

1 answer:

Lynna [10]4 years ago

5 0

Answer:

a) 60% probability that he picks up a blue pen and recognizes it as blue.

b) 62.5% probability that he chooses a pen and thinks it is blue.

c) 96% probability that he actually chose a blue pen.

Step-by-step explanation:

The problem has the following probabilities:

75 of 100 pens are blue, so a 75% probability that a pen is blue.

25 0f 100 pens are green, so a 25% probability that a pen is green.

If a pen is blue, an 80% probability that he thinks it is a blue pen

If a pen is blue, a 20% probability that he thinks it is a green pen.

If a pen is green, a 90% probability that he thinks it is a green pen.

If a pen is green, a 10% probability that that he thinks it is a blue pen.

a) (3pts) What is the probability that he picks up a blue pen and recognizes it as blue?

75% probability he picks up a blue pen.

80% that he recognizes a blue pen as blue.

$P = 0.75*0.8 = 0.6$

60% probability that he picks up a blue pen and recognizes it as blue.

b) (3 pts) What is the probability that he chooses a pen and thinks it is blue?

75% probability that he picks up a blue pen.

80% that he recognizes a blue pen as blue.

25% probability that he picks up a green pen

10% probability that he thinks a green pen is blue.

$P = 0.75*0.80 + 0.25*0.10 = 0.625$

62.5% probability that he chooses a pen and thinks it is blue.

c) (4pts) Given that he thinks he chose a blue pen, what is the probability that he actually chose a blue pen?

62.5% probability that he thinks that he choose a blue pen.

60% probability that he chooses a blue pen and think that it is blue.

$P = \frac{0.6}{0.625} = 0.96$

96% probability that he actually chose a blue pen.

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4 years ago

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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:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

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

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

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This means that $p = \frac{99}{99+78} = 0.5593$

Mean and standard deviation:

$\mu = p = 0.5593$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.5593*0.4407}{99+78}} = 0.0373$

a. Calculate the probability of getting more than 61% green balls.

This is 1 subtracted by the pvalue of Z when X = 0.61. So

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

By the Central Limit Theorem

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

$Z = \frac{0.61 - 0.5593}{0.0373}$

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$Z = 1.36$ has a pvalue of 0.9131

1 - 0.9131 = 0.0869

0.0869 = 8.69% probability of getting more than 61% green balls.

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