One barge from Inland Waterways, Inc. can carry a load of 2555 lb. Records of past trips show that

the weights of the cans that it carries have a mean of 98 lb and a standard deviation of 10 lb. For

samples of size 25, find the mean and standard deviation of x.

A) x = 98; x = 2

B) x = 98; x = 10

C) x = 2; x = 98

D) x = 10; x = 98

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

[tex]\mu = 98, \sigma = 10[/tex]

Samples of size 25

Mean 98

Standard deviation [tex]s = \frac{10}{\sqrt{25}} = 2[/tex]

So the correct answer is:

A) x = 98; x = 2

Cricetus Cricetus · Answer 2 · 2021-12-20T15:41:37+01:00

The mean and standard deviation will be "x = 98; x = 2".

Given values:

Sample mean,

Population parameter, [tex]\mu = 98[/tex]

Sample of size,

25

and,

[tex]\sigma = 10[/tex]

Now,

The sample standard deviation will be:

= [tex]\frac{\sigma}{\sqrt{n} }[/tex]

By putting the values, we get

= [tex]\frac{10}{\sqrt{25} }[/tex]

= [tex]\frac{10}{5}[/tex]

= [tex]2[/tex]

Thus the above choice i.e., "option A" is appropriate.

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