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3 years ago

Weights measured in grams of randomly selected M&M plain candies:

Mathematics

1 answer:

charle [14.2K]3 years ago

6 0

Answer:

The range of weights of the middle 99.7% of M&M’s is between 0.8187 and 1.0203.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

Sample mean:

$\overline{x} = \frac{0.957 + 0.912 + 0.925 + 0.886 + 0.920 + 0.958 + 0.915 + 0.914 + 0.947 + 0.939 + 0.842}{11} = 0.9195$

Sample standard deviation:

$s = \sqrt{\frac{(0.957-0.9195)^2 + (0.912-0.9195)^2 + (0.925-0.9195)^2 + (0.886-0.9195)^2 + ...}{10}} = 0.0336$

What is the range of weights of the middle 99.7% of M&M’s?

By the Empirical Rule, within 3 standard deviations of the mean, so:

0.9195 - 3*0.0336 = 0.8187.

0.9195 + 3*0.0336 = 1.0203.

The range of weights of the middle 99.7% of M&M’s is between 0.8187 and 1.0203.

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Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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$\sqrt{V(X)} = \sqrt{np(1-p)}$

9% of all drivers arrested for DUI (Driving Under the Influence) are repeat offenders

This means that $p = 0.09$

41 people arrested for DUI in Illinois are selected at random.

This means that $n = 41$

a. What is the probability that exactly 3 people are repeat offenders?

This is P(X = 3).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{41,3}.(0.09)^{3}.(0.91)^{38} = 0.2158$

21.58% probability that exactly 3 people are repeat offenders

b. What is the probability that at least one person is a repeat offender?

Either none are repeat offenders, or at least one is. The sum of the probabilities of these outcomes is 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$ .

Then

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = 0) = C_{41,0}.(0.09)^{0}.(0.91)^{41} = 0.0209$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0209 = 0.9791$

97.91% probability that at least one person is a repeat offender

c. What is the mean number of repeat offenders?

$E(X) = np = 41*0.09 = 3.69$

d. What is the standard deviation of the number of repeat offenders?

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{41*0.09*0.91} = 1.83$

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