The amounts due on a mobile phone bill in Ireland are normally distributed with a mean of € 53 and a standard deviation of € 15. If a monthly phone bill is chosen at random, find the probability that we get a phone bill between 47 and 74 euros.

Answer:0.5746Step-by-step explanation:Lets call X monthly bill chosen. X is a random variable with distribution N(μ=53,σ =15). We first standarize X, lets call W = (X-53)/15, random variable obtained from X by substracting μ and dividing by σ. W is a random variable with distribution N(0,1) and the cummulative function of W, Φ, is tabulated. You can find the values of Φ on the attached file. Using this information we obtain[tex] P(47 < X < 74) = P(\frac{47-53}{15} < \frac{X-53}{15} < \frac{74-53}{15}) = P(-0.4 < W < 1.4) = \phi(1.4) - \phi(-0.4) [/tex]A standard normal random variable has a symmetric density function, as a result Φ(-0.4) = 1- Φ(0.4) = 1- 0.6554 = 0.3446. Also, Φ(1.4) = 0.9192. We conclude that Φ(1.4)-Φ(-0.4) =  0.9192-0.3446 = 0.5746, that is the probability that the phone bill is between 47 and 74 euros.