We have a normally distributed variable X with mean 33 minutes and standard deviation of 12 minutes.
a. The distribution of X can be expressed as:
[tex]\begin{gathered} \mu=33 \\ \sigma^2=12^2=144 \\ \Rightarrow X\sim N(33,144) \end{gathered}[/tex]b. We can calculate the probability P(X > 37) using the z-score for X = 37 and then looking for the standard normal probability for that z-score.
We can calculate the z-score as:
[tex]\begin{gathered} z=\frac{X-\mu}{\sigma} \\ z=\frac{37-33}{12} \\ z=\frac{4}{12} \\ z=\frac{1}{3} \end{gathered}[/tex]We can now look for the probability as:
[tex]P(X>37)=P(z>\frac{1}{3})=0.3694[/tex]c. The 80th percentile is the value of X for which 80% of the data is below that value.
We can find the z-score that represents the 80th percentile in the standard normal distribution and then convert it to the equivalent value of X in the actual distribution.
Then, we start finding z:
[tex]\begin{gathered} P(zThen, we can zonvert this value of z = 0.84162 into an equivalent value of X as:[tex]\begin{gathered} X_{80}=\mu+z_{80}\cdot\sigma \\ X_{80}=33+0.84162\cdot12 \\ X_{80}=33+10.09944 \\ X_{80}\approx43.0994 \end{gathered}[/tex]Answer:
a. X ~ N(33, 144)
b. The probability is 0.3694.
c. The 80th percentile is P80 = 43.0994.
