Question

Americans spend an average of 2 hours per day online. If the standard deviation is 23 minutes, find the range in which at least 75% of the data will lie.

Answer 1

Answer:

75% of the data lies on the interval [1.5592, 2.4408]

Step-by-step explanation:

Let X be the amount of hours a random american spends online per day. X has normal distribution with parameters μ = 2 and σ = 23/60 = 0.383 . The standarization of X, W is

$W = \frac{X-\mu}{\sigma} = \frac{X-2}{0.383}$

Which has distribution N(0,1). The values of the cummulative distribution function of W are tabulated and they can be found in the attached file.

We will find and interval centered in the mean 2 in which the probability that a random american belongns to that interval is 0.75. The interval has the form

I = [2-Z , 2+Z]

As a consecuence

$0.75 = P(X \in I) = P(X \in [2-Z\,,\,2+Z ] ) = P(2-Z < X < 2+Z) = \\P(\frac{2-Z-2}{0.383} < W < \frac{2+Z-2}{0.383}) = P(- \frac{Z}{0.383} < W < \frac{Z}{0.383}) = \phi(\frac{Z}{0.383}) - \phi(- \frac{Z}{0.383})$

Since W is symmetric respect to 0, then $\phi(- \frac{Z}{0.383}) = 1-\phi(\frac{Z}{0.383}).$

Thus, the last expression is equal to $2 \phi(\frac{Z}{0.383}) - 1$ . We conclude that

$0.75 = 2 \phi(\frac{Z}{0.383}) - 1 \Rightarrow \phi(\frac{Z}{0.383}) = 0.875$

Looking at the table, we will find that Z/0.383 = 1.15, then Z = 1.15*0.383 = 0.4408. Therefore, the interval we are looking for is [2-0.4408, 2+0.4408] = [1.5592, 2.4408].