Question

The distribution of weights of potato chip bags filled off a production line is unknown. However, the mean is m=13.35 OZs and the standard deviation is s=.1200 OZs. To check the quality of the potato chip bags, a random sample of n=36 bags is selected. Use `X to denote the sample mean.A) The mean (or expected value) of the sample mean `X isa.13.23b.13.00c.13.47d.13.50e.none of the aboveB) The standard deviation of the sample mean `X isa.0.0200b.0.1200c.0.0600d.0.3600e.none of the aboveC) The probability that the sample mean `X is less than or equal to 13.38 OZs is closest toa.0.0668b.0.0998c.0.4332d.0.9332e.none of the aboveD) The probability that the sample mean `X is less than or equal to 13.32 OZs is closest toa.0.0668b.0.0998c.0.4332d.0.9332e.none of the aboveE) The probability that the sample mean `X is between 13.30 and 13.36 OZs is closest toa.0.0062b.0.3085c.0.6853d.0.4938e.none of the above

Answer 1

a)
`\mu_(\bar X) =13.35`

e.none of the above

b)
`\sigma_(\bar X)= (0.12)/(√(36))= 0.02`

a.0.0200

c)
`z = (13.32-13.35)/((0.12)/(√(36)))= -1.5`

`P(\bar X< 13.32)= P(z<-1.5)=0.0668`

a.0.0668

d)
`z = (13.30-13.35)/((0.12)/(√(36)))= -2.5`

`z = (13.36-13.35)/((0.12)/(√(36)))= 0.5`

`P(13.30<\bar X< 13.35)= P(-2.5<z<0.5)=P(z<0.5) -P(z<-2.5) =0.69146-0.00621=0.6853`

c.0.6853

Explanation:

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".

Solution to the problem

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(13.35,0.12)`

Where
`\mu=13.35` and
`\sigma=0.12`

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)))`

Part a

`\mu_(\bar X) =13.35`

Part b

`\sigma_(\bar X)= (0.12)/(√(36))= 0.02`

Part c

We want this probability:

`P(\bar X< 13.32)`

For this case we can use the z score formula given by:

`z = (\bar X -\mu)/((\sigma)/(√(n)))`

And using this formula we got:

`z = (13.32-13.35)/((0.12)/(√(36)))= -1.5`

`P(\bar X< 13.32)= P(z<-1.5)=0.0668`

Part d

We want this probability:

`P(13.30<\bar X< 13.36)`

For this case we can use the z score formula given by:

`z = (\bar X -\mu)/((\sigma)/(√(n)))`

And using this formula we got:

`z = (13.30-13.35)/((0.12)/(√(36)))= -2.5`

`z = (13.36-13.35)/((0.12)/(√(36)))= 0.5`

`P(13.30<\bar X< 13.35)= P(-2.5<z<0.5)=P(z<0.5) -P(z<-2.5) =0.69146-0.00621=0.6853`