Question

The times taken by students to get up in the morning can be modelled by a normal distribution with
mean 26.4 minutes and standard deviation 3.7 minutes.

‘Very slow’ students are students whose time to get up is more than 1.645 standard deviations
above the mean. Find the probability that fewer than 3 students from a random sample of 8
students are ‘very slow’.

Answer 1

This is a binomial probability in disguise. Given that the times are normally distributed, you can use this information to find the probability that a random student picked from a population is considered "very slow."

You're told that "very slow" students are those whose time is greater than 1.645 std devs from the mean, which means \(P(Z>1.645)=0.05=5\%\). This will be your success probability, where "success" refers to the probability that a student is "very slow".

Now, from a sample of 8 students \((n=8)\), you want to find the probability that less than 3 students are "very slow". If we call the random variable for this binomial distribution \(T\), we have
\[\begin{align*}P(T<3)&=P(T=0)+P(T=1)+P(T=2)\\
&=\binom80(0.05)^0(0.95)^{8-0}+\binom81(0.05)^1(0.95)^{8-1}+\binom82(0.05)^2(0.95)^{8-2}\\

\end{align*}\]