Mathematics, 16.07.2021 02:50 ctyrector

A packing plant fills bags with cement. The weight X kg of a bag can be modeled by a normal distribution with mean 50kg and standard deviation 2kg. 4. a. Find the probability that a randomly selected bag weighs more than 53kg.
b. Find the weight that is exceeded by 98% of the bags.
c. Three bags are selected at random. Find the probability that two weigh more than 53kg and one weighs less than 53kg.

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Answer from: trevorhenyan51

a) 0.0668 = 6.68% probability that a randomly selected bag weighs more than 53kg.

b) The weight that is exceeded by 98% of the bags is of 45.9 kg.

c) 0.0125 = 1.25% probability that two weigh more than 53kg and one weighs less than 53kg.

Step-by-step explanation:

The first two questions are solved using the normal distribution, while the third is solved using the binomial distribution.

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.

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.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The weight X kg of a bag can be modeled by a normal distribution with mean 50kg and standard deviation 2kg.

This means that $\mu = 50, \sigma = 2$

a. Find the probability that a randomly selected bag weighs more than 53kg.

This is 1 subtracted by the p-value of Z when X = 53. So

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

$Z = \frac{53 - 50}{2}$

Z = 1.5

Z = 1.5 has a p-value of 0.9332.

1 - 0.9332 = 0.0668.

0.0668 = 6.68% probability that a randomly selected bag weighs more than 53kg.

b. Find the weight that is exceeded by 98% of the bags.

This is the 100 - 98 = 2nd percentile, which is X when Z has a p-value of 0.02, so X when Z = -2.054.

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

$-2.054 = \frac{X - 50}{2}$

X - 50 = -2.054*2

X = 45.9

The weight that is exceeded by 98% of the bags is of 45.9 kg.

c. Three bags are selected at random. Find the probability that two weigh more than 53kg and one weighs less than 53kg.

0.0668 = 6.68% probability that a randomly selected bag weighs more than 53kg means that p = 0.0668

3 bags means that n = 2

Two above 53kg, which means that we want P(X = 2). So

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

$P(X = 2) = C_{3,2}.(0.0668)^{2}.(0.9332)^{1} = 0.0125$

0.0125 = 1.25% probability that two weigh more than 53kg and one weighs less than 53kg.

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