Question

Suppose certain coins have weights that are normally distributed with a mean of 5.805 g5.805 g and a standard deviation of 0.071 g0.071 g. A vending machine is configured to accept those coins with weights between 5.6755.675 g and 5.9355.935 g. a. If 260260 different coins are inserted into the vending​ machine, what is the expected number of rejected​ coins?

Answer 1

a)
`P(5.675<X<5.935)=P(-1.83<Z<1.83)=P(Z<1.83)-P(Z<-1.83)=0.9664-0.0336=0.933`

And the expected number would be 260*0.933=242.58 and n =243 rounded up.

b)
`P(5.675< \bar X <5.935)=P((5.675-5.805)/((0.071)/(√(260)))<Z<(5.935-5.805)/((0.071)/(√(260))))=P(-29.524<Z<29.524) = 1`

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

2) Part a

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

`X \sim N(5.805,0.71)`

Where
`\mu=5.805` and
`\sigma=0.071`

We are interested on this probability

`P(5.675<X<5.935)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/(\sigma)`

If we apply this formula to our probability we got this:

`P(5.675<X<5.935)=P((5.675-\mu)/(\sigma)<(X-\mu)/(\sigma)<(5.935-\mu)/(\sigma))=P((5.675-5.805)/(0.071)<Z<(5.935-5.805)/(0.071))=P(-1.83<Z<1.83)`

And we can find this probability on this way:

`P(-1.83<Z<1.83)=P(Z<1.83)-P(Z<-1.83)=0.9664-0.0336=0.933`

And the expected number would be 260*0.933=242.58 and n =243 rounded up.

3) Part b

If 260 different coins are inserted in the vending machine, what is the probability that the mean falls between the limits of 5.675 g and 5.935.

From the central limit theorem we know that the distribution for the sample mean
`\bar X` is given by:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

`P(5.675< \bar X <5.935)=P((5.675-5.805)/((0.071)/(√(260)))<Z<(5.935-5.805)/((0.071)/(√(260))))=P(-29.524<Z<29.524) = 1`