Question

The amount of cola in a 355 ml bottle from a certain company is a random variable with a mean of 355 ml and a standard deviation of 2 ml. For a sample of size 32, perform the following calculations. a. Find an approximate probability that the sample mean is less than 354.8 ml. b. Suppose the amount of cola is distributed as N(355, 4). Find an approximate probability that 10 of the bottles in the sample contain less than 354.8 ml of cola.

Answer 1

The probability of the sample mean being less than 354.8 ml can be calculated using the z-score and the standard normal distribution. For the second part, if the standard deviation is indeed 4, it's necessary to calculate the probability for a single bottle first, then apply the binomial distribution to find the probability for 10 out of 32 bottles.

Step-by-step explanation:

The calculation of the probability that the sample mean is less than 354.8 ml when the population mean is 355 ml and the standard deviation is 2 ml for a sample size of 32 uses the standard normal distribution. Applying the Central Limit Theorem, we can find the z-score and use it to calculate the probability.

For part b, assuming the normal distribution N(355, 4), to find the probability that 10 out of 32 bottles contain less than 354.8 ml, we would use the binomial distribution with the probability calculated in part a. However, the second part of question b appears to have a typo with the standard deviation. If the distribution is N(355, 4), we would need to first find the probability of a single bottle having less than 354.8 ml, then use the binomial formula to find the probability of 10 bottles having less than that amount.

Answer 2

a)
`P(\bar X<354.8)=0.284`

b)
`P(\bar X<354.8)=0.436`

Step-by-step 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".

Let X the random variable that represent the amount of cola in a 355 ml bottle from a certain company, and for this case we know the distribution for X is given by:

`X \sim N(\mu=355,\sigma=2)`

And let
`\bar X` represent the sample mean, the distribution for the sample mean is given by:

`\bar X \sim N(\mu,(\sigma)/(√(n)))`

On this case
`\bar X \sim N(355,(2)/(√(32)))`

Part a

We want this probability:

`P(\bar X<354.8)=P(\bar X<354.8)`

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

`z=(x-\mu)/((\sigma)/(√(n)))`

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

`P(\bar X <354.8)=P(Z<(354.8-355)/((2)/(√(32))))=P(Z<-0.57)`

`P(\bar X<354.8)=0.284`

Part b

Let X the random variable that represent the amount of cola in a 355 ml bottle from a certain company, and for this case we know the distribution for X is given by:

`X \sim N(\mu=355,\sigma=4)`

And let
`\bar X` represent the sample mean, the distribution for the sample mean is given by:

`\bar X \sim N(\mu,(\sigma)/(√(n)))`

On this case
`\bar X \sim N(355,(4)/(√(10)))`

We want this probability:

`P(\bar X<354.8)=P(\bar X<354.8)`

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

`z=(x-\mu)/((\sigma)/(√(n)))`

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

`P(\bar X <354.8)=P(Z<(354.8-355)/((4)/(√(10))))=P(Z<-0.16)`

`P(\bar X<354.8)=0.436`