Question

The amount of time spent by adults reading the New York Times online can be seen as normally distributed with a mean of 15 minutes and a standard deviation of 10 minutes.

1. The probability of reading for less than 18 minutes is

2. The probability of reading for less than 18 minutes is

3. 15% of readers spend more than how long?

4. What is the probability of spending less than no time reading? Does that make sense? Explain why it makes sense or not.

Answer 1

Let X be the amount of time spent by adults reading the New York Times online. Then X~N(15,10²). We will use the result

Z = where is the mean and is the standard deviation. We will obtain P(Z<z) values using probabilities of the standard normal distribution

1)  The probability of reading for less than 18 minutes i.e P(X <18) = P(Z < (18-15)/10) = P(Z<0.3) = 0.6179

2) The probability of reading for more than 18 minutes i.e P(X>18) = 1- P(X<18) = 1-0.6179 = 0.3821

3) Let a be the time that 15% of readers spend more than i.e P (X>a) = 0.15 or P(Z > (a-15)/10) = 0.15 => (a-15)/10 = 1.0364

=> a=1.0364*10 +15 => a= 25.36

Hence 15% readers spend more than 25.36 minutes reading.

4)  the probability of spending less than no time reading i.e P(X<0) = P(Z< -1.5) = P(Z>1.5) = 1-P(Z<1.5) = 1-0.9332 = 0.0668

Since this is a continuous distribution, it makes sense to calculate this probability as P(X=0) = 0. The probability P(X<0) represents the probability of not reading at all and is the only way to calculate such probability.