Question

Assume that the weight of hamsters is Normally distributed with μ = 1.2 lbs and σ = 0.25 lbs. Suppose I randomly select 4 hamsters.

(a) What is the shape of the sampling distribution of the sample mean? (unknown, exactly normal, approximately normal)

(b) What is P(X < 1.14)? (Use 3 decimal places)

Answer 1

Given Information:

Population mean = μ = 1.2

Population standard deviation = σ = 0.25

Sample size n = 4

Required Information:

P(X < 1.14) = ?

Answer:

The probability of P(X < 1.14) is 0.315

Explanation:

(a) What is the shape of the sampling distribution of the sample mean?

Since it is given that population is normally distributed then sample distribution will also be normally distributed even with a smaller sample size .

When population is not normally distributed and is very skewed then a larger sample size is required to ensure normal sample distribution according to central limit theorem.

(b) What is P(X < 1.14)?

First let us find out sample mean and standard deviation

sample mean will be same as population mean μ = 1.2

sample standard deviation = σ/√n

where n = 4 is given

sample standard deviation = σ/√4 = 0.125

Now we can calculate the probability using z-score

P(X < 1.14) = P(Z < (1.14 - 1.2)/0.125) = P(Z < -0.48)

From the z-table

P(Z < -0.48) = 0.3156

Therefore, the probability of P(X < 1.14) is 0.315 or 31.5%

Answer 2

a)
`\bar X \sim N(\mu, (\sigma)/(√(n)))`

With the following parameters:

`\mu_(\bar X)= \mu = 1.2`

`\sigma_(\bar X) = (0.25)/(√(4))= 0.125`

b)
`P(\bar X <1.14)`

And we can use the z score given by:

`z =(\bar X -\mu)/((\sigma)/(√(n)))`

Using this formula we got:

`P(\bar X <1.14)=P(Z<(1.14-1.2)/((0.25)/(√(4)))=-0.48)`

And using a calculator, excel or the normal standard table we have that:

`P(Z<-0.48)=0.316`

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".

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(1.2,0.25)`

Where
`\mu=1.2` and
`\sigma=0.25`

And we select a random sample of n=4

Since the distribution for X is normal then we can conclude that the distribution for the sample mean
`\bar X` is given by:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

With the following parameters:

`\mu_(\bar X)= \mu = 1.2`

`\sigma_(\bar X) = (0.25)/(√(4))= 0.125`

Part b

We want this probability:

`P(\bar X <1.14)`

And we can use the z score given by:

`z =(\bar X -\mu)/((\sigma)/(√(n)))`

Using this formula we got:

`P(\bar X <1.14)=P(Z<(1.14-1.2)/((0.25)/(√(4)))=-0.48)`

And using a calculator, excel or the normal standard table we have that:

`P(Z<-0.48)=0.316`