Part a.
From the given infomation, the mean is equal to
[tex]\mu=2.9\text{ hours}[/tex]
and the standard deviation
[tex]\sigma=1.4\text{ hours}[/tex]
Then, the distribution of X is:
[tex]N(2.9,1.4)[/tex]
Part b.
In this case, we need to find the following probability:
[tex]P(X<2.6)[/tex]
So, in order to find this value, we need to convert the 2.6 hours into a z-value score by means of the z-score formula:
[tex]z=\frac{X-\mu}{\sigma}[/tex]
Then, by substituting the given values into the formula, we get
[tex]\begin{gathered} z=\frac{2.6-2.9}{1.4} \\ z=-0.214285 \end{gathered}[/tex]
Then, the probability we must find in the z-table is:
[tex]P(z<-0.214285)[/tex]
which gives
[tex]P(z<-0.214285)=0.41516[/tex]
Therefore, by rounding to 4 decimal places, the answer for part b is: 0.4152
Part c.
In this case, we need to find the following probability
[tex]P(X>2.5)[/tex]
Then, by converting 2.5 to a z-value, we have
[tex]\begin{gathered} z=\frac{2.5-2.9}{1.4} \\ z=-0.285714 \end{gathered}[/tex]
So, we need to find on the z-table:
[tex]P(z>-0.285714)[/tex]
which gives
[tex]P(z\gt-0.285714)=0.61245[/tex]
Then, by multiplying this probability by 100% and rounding to the nearest hundreadth,
the answer for part c is: 61.25 %
Part d.
In this case, we have the following information:
[tex]P(z>Z)=0.72[/tex]
and we need to find Z. From the z-table, we get
[tex]Z=0.58284[/tex]
Then, from the z-value formula, we have
[tex]-0.58284=\frac{X-2.9}{1.4}[/tex]
and we need to isolate the amount of hours given by X. Then, by multiplying both sides by 1.4, we obtain
[tex]-0.815976=X-2.9[/tex]
Then, X is given by
[tex]\begin{gathered} X=2.9-0.815976 \\ X=2.0840 \end{gathered}[/tex]
So, by rounding to 4 decimal places, the answer is: 2.0840 hours.
