The time that John needs to get from his house to the University every morning is a continuous random variable which is assumed to follow the normal distribution with mean 𝜇 = 35 minutes and a standard deviation 𝜎 = 5 minutes.(i) Find the probability that on a particular day his journey takes (a)

Question

The time that John needs to get from his house to the University every morning is a continuous random variable which is assumed to follow the normal distribution with mean 𝜇 = 35 minutes and a standard deviation 𝜎 = 5 minutes.

(i) Find the probability that on a particular day his journey takes
(a) less than 30 minutes;
(b) between 30 and 40 minutes.

(ii) Tomorrow, John’s first lecture starts at 10:15 a.m. and he does not want to be late. Estimate what time he should leave his house so that he arrives at the classroom before the lecture starts with probability 99%.

Accepted Answer

Let X be the time that John’s journey takes on that day. Then, it is given that X ∼ N(35, 5²).

(i) (a) The required probability is P(X < 30) which is

[latex]P(X\lt 30)=P\left({\frac{X-\mu}{\sigma}}\lt {\frac{30-\mu}{\sigma}}\right)=P\left({\frac{X-35}{5}}\lt {\frac{30-35}{5}}\right)=P\left(Z\lt {\frac{-5}{5}}\right)=P(Z\lt -1)=\Phi(-1)=1-\Phi(1),[/latex]

wherein we have used Proposition 7.4 in the last step (recall that the table in Appendix B gives Φ(z) only for nonnegative z). From this table, we see that Φ(1) = 0.8413, and so the required result is 1 − 0.8413 = 0.1587, i.e. a probability of about 16%.

(b) The probability we seek for this part is P(30 ≤ X ≤ 40), which is

[latex]P(30\leq X\leq40)=P\left({\frac{30-\mu}{\sigma}}\leq{\frac{X-\mu}{\sigma}}\leq{\frac{40-\mu}{\sigma}}\right)=P\left({\frac{30-35}{5}}\leq{\frac{X-35}{5}}\leq{\frac{40-35}{5}}\right)=P(-1\leq Z\leq1),[/latex]

and we have already seen in (7.9)

[latex]P(−1 \leq Z \leq 1) = 2(0.8413) − 1 = 0.6826 \cong 68%,[/latex] (7.9)

that this equals 0.6826. Therefore, the probability that John’s journey to the University will take between 30 and 40 minutes is about 68%.

(ii) Now the problem is of a different type. Specifically, we want to find the value of x such that the journey time, X, is at most x with probability 99%, i.e.
[latex] P(X \leq x) = 0.99.[/latex] (7.12)

For this, we express the probability P(X ≤ x) in terms of the standard normal distribution, working in a familiar way, as follows:

[latex]P(X\leq x)=P\left({\frac{X-\mu}{\sigma}}\leq{\frac{x-\mu}{\sigma}}\right)=P\left(Z\leq{\frac{x-\mu}{\sigma}}\right)=\Phi\left({\frac{x-\mu}{\sigma}}\right).[/latex]

If we put for simplicity [latex]z=(x-\mu)/\sigma,[/latex] we then want to find the value of z for which

[latex]\Phi(z)=0.99.[/latex]

So, in this case, we are given the (cumulative) probability P(Z ≤ z) and we seek the value of z. From the standard normal table in Appendix B, we see that this value is z = 2.33. Since the value of x which satisfies (7.12) is given by x = 𝜇 + 𝜎z, for z = 2.33 we finally obtain

[latex]x=\mu+\sigma z=35+5\cdot2.33=46.65\cong47\,{\mathrm{minutes}}.[/latex]

This is the maximum journey time that John should allow himself in order to arrive in time for the lecture with probability 99%. Since the lecture starts at 10:15 a.m., this means that he should leave his house by 9:28 a.m.

Question 7.6: The time that John needs to get from his house to the Univer......

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