The daily exchange rates for the five-year period 2003 to 2008 between the euro (EUR) and the British pound (GBP) are well-modeled by a normal distribution with mean 1.459 euros (to pounds), and standard deviation 0.033 euros. What is the probability that on a randomly selected day during this period, the pound was worth (1) less than 1.459 euros, (2) more than 1.492 euros

When trying to solve this, I did:

1) z=(1.495-1.495)/0.033 and then matched the answer of 0 with z<0 on the standard normal probabilities table to get 0.5, which is correct.

2)(1.492-1.495)/0.033 and got the answer 0.0909, with which I matched it with 0.09 on the table and got 0.4641, but the correct answer is 0.1587 and I am confused about how to get this answer.

Any help is much appreciated!

  • $\begingroup$ Let $X$ be the random variable of the exchange rate. In the second question, you are looking for $P(X \geq 1.492) = 1 - P(X \leq 1.492)$. Did you remember to do this step (subtract from 1)? $\endgroup$
    – cgo
    Nov 6, 2015 at 1:38

1 Answer 1


Let the random variable $X=$The price of the pound in euros on a randomly selected day within this period.


$$X\sim N\left(1.459,0.033^2\right)$$ $$P(X \ge 1.492)=P\left(\cfrac{1.459 - 99}{0.033} \le X \le \cfrac{1.459-1.492}{0.033}\right)=P(-2955 \le z \le -1)=\color{red}{\fbox{0.158655253}}$$ as required.

If you are wondering where the $99$ came from I simply selected an arbitrarily large value to ensure all values of $X$ were covered up to $X=1.492$.

As you are using tables you can simply select a very large and negative $z$ value.

  • $\begingroup$ thank you very much for the help! I understand now. $\endgroup$
    – Astag
    Nov 6, 2015 at 2:55
  • $\begingroup$ I just have one question, I thought that in order to find the z-score you must input the values into the formula z=(x-μ)/σ... why is it not (1.492-1.459)/0.033? Once again, thank you for the help. $\endgroup$
    – Astag
    Nov 6, 2015 at 13:42
  • 1
    $\begingroup$ @Astag Yes the formula is $\cfrac{x-\mu}{\sigma}$ but switching the order of $x$ and $\sigma$ only changes the sign, so the way I have done it will still give the correct answer $\endgroup$
    – BLAZE
    Nov 6, 2015 at 14:19
  • $\begingroup$ Oh okay, I just thought that the final answer would vary based on the sign. $\endgroup$
    – Astag
    Nov 6, 2015 at 14:22
  • 1
    $\begingroup$ Thank you for clearing this up, it is much appreciated. $\endgroup$
    – Astag
    Nov 6, 2015 at 14:30

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