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.

APPENDIX B DISTRIBUTION FUNCTION OF THE STANDARD NORMAL DISTRIBUTION
$\Phi(z)={\frac{1}{\sqrt{2\pi}}}\int_{- \infty}^{z}e^{-x^2/2}\,\mathrm{d}x$ $\Phi(-z) = 1 – \Phi(z)$
z	0.00	0.01	0.02	0.03	0.04	0.05	0.06	0.07	0.08	0.09
0.0	0.5000	0.5040	0.5080	0.5120	0.5160	0.5199	0.5239	0.5279	0.5319	0.5359
0.1	0.5398	0.5438	0.5478	0.5517	0.5557	0.5596	0.5636	0.5675	0.5714	0.5753
0.2	0.5793	0.5832	0.5871	0.5910	0.5948	0.5987	0.6026	0.6064	0.6103	0.6141
0.3	0.6179	0.6217	0.6255	0.6293	0.6331	0.6368	0.6406	0.6443	0.6480	0.6517
0.4	0.6554	0.6591	0.6628	0.6664	0.6700	0.6736	0.6772	0.6808	0.6844	0.6879
0.5	0.6915	0.6950	0.6985	0.7019	0.7054	0.7088	0.7123	0.7157	0.7190	0.7224
0.6	0.7257	0.7291	0.7324	0.7357	0.7389	0.7422	0.7454	0.7486	0.7517	0.7549
0.7	0.7580	0.7611	0.7642	0.7673	0.7703	0.7734	0.7764	0.7794	0.7823	0.7852
0.8	0.7881	0.7910	0.7939	0.7967	0.70.995	0.8023	0.8051	0.8078	0.8106	0.8133
0.9	0.8159	0.8186	0.8212	0.8238	0.8264	0.8289	0.8315	0.8340	0.8365	0.8389
1.0	0.8413	0.8438	0.8461	0.8485	0.8508	0.8531	0.8554	0.8577	0.8599	0.8621
1.1	0.8643	0.8665	0.8686	0.8708	0.8729	0.8749	0.8770	0.8790	0.8810	0.8830
1.2	0.8849	0.8869	0.8888	0.8907	0.8925	0.8944	0.8962	0.8980	0.8997	0.9015
1.3	0.9032	0.9049	0.9066	0.9082	0.9099	0.9115	0.9131	0.9147	0.9162	0.9177
1.4	0.9192	0.9207	0.9222	0.9236	0.9251	0.9265	0.9279	0.9292	0.9306	0.9319
1.5	0.9332	0.9345	0.9357	0.9370	0.9382	0.9394	0.9406	0.9418	0.9429	0.9441
1.6	0.9452	0.9463	0.9474	0.9484	0.9495	0.9505	0.9515	0.9525	0.9535	0.9545
1.7	0.9554	0.9564	0.9573	0.9582	0.9591	0.9599	0.9608	0.9616	0.9625	0.9633
1.8	0.9641	0.9649	0.9656	0.9664	0.9671	0.9678	0.9686	0.9693	0.9699	0.9706
1.9	0.9713	0.9719	0.9726	0.9732	0.9738	0.9744	0.9750	0.9756	0.9761	0.9767
2.0	0.9772	0.9778	0.9783	0.9788	0.9793	0.9798	0.9803	0.9808	0.9812	0.9817
2.1	0.9821	0.9826	0.9830	0.9834	0.9838	0.9842	0.9846	0.9850	0.9854	0.9857
2.2	0.9861	0.9864	0.9868	0.9871	0.9875	0.9878	0.9881	0.9884	0.9887	0.9890
2.3	0.9893	0.9896	0.9898	0.9901	0.9904	0.9906	0.9909	0.9911	0.9913	0.9916
2.4	0.9918	0.9920	0.9922	0.9925	0.9927	0.9929	0.9931	0.9932	0.9934	0.9936
2.5	0.9938	0.9940	0.9941	0.9943	0.9945	0.9946	0.9948	0.9949	0.9951	0.9952
2.6	0.9953	0.9955	0.9956	0.9957	0.9959	0.9960	0.9961	0.9962	0.9963	0.9964
2.7	0.9965	0.9966	0.9967	0.9968	0.9969	0.9970	0.9971	0.9972	0.9973	0.9974
2.8	0.9974	0.9975	0.9976	0.9977	0.9977	0.9978	0.9979	0.9979	0.9980	0.9981
2.9	0.9981	0.9982	0.9982	0.9983	0.9984	0.9984	0.9985	0.9985	0.9986	0.9986
3.0	0.9987	0.9987	0.9987	0.9988	0.9988	0.9989	0.9889	0.9889	0.9890	0.9990
3.1	0.9990	0.9991	0.9991	0.9991	0.9992	0.9992	0.9992	0.9992	0.9993	0.9993
3.2	0.9993	0.9993	0.9994	0.9994	0.9994	0.9994	0.9994	0.9995	0.9995	0.9995
3.3	0.9995	0.9995	0.9995	0.9996	0.9996	0.9996	0.9996	0.9996	0.9996	0.9997
3.4	0.9997	0.9997	0.9997	0.9997	0.9997	0.9997	0.9997	0.9997	0.9997	0.9998
3.5	0.9998	0.9998	0.9998	0.9998	0.9998	0.9998	0.9998	0.9998	0.9998	0.9998

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

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