Of all the companies on the New York Stock Exchange, profits are normally distributed with a mean of $6.54 million and a standard deviation of $10.45 million. In a random sample of 73 companies from the NYSE, what is the probability that the mean profit for the sample was between 3.0 million and 6.0 million?

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".

Solution to the problem

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

[tex]X \sim N(6.54,10.45)[/tex]

Where [tex]\mu=6.54[/tex] and [tex]\sigma=10.45[/tex]

We are interested on this probability :

[tex]P(3.0<\bar X<6.0)[/tex]

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

[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

And we can find the z score for the two limits 3.0 and 6.0 and we got:

[tex] z_1 = \frac{3-6.54}{\frac{10.45}{\sqrt{73}}} = -2.894[/tex]

[tex] z_2 = \frac{6- 6.54}{\frac{10.45}{\sqrt{73}}} =-0.442 [/tex]

So then we can calculate the probability on this way:

[tex]P(3.0<\bar X<6.0)= P(-2.894< Z< -0.442) = P(Z<-0.442) -P(Z<-2.894)[/tex]

And we can use the normal standard distribution or excel to calculate the probabilities and we got:

[tex]P(3.0<\bar X<6.0)= P(-2.894< Z< -0.442) = P(Z<-0.442) -P(Z<-2.894) = 0.329 -0.0019=0.327[/tex]