Question

according to a study by rescuetime, people spend an average of 1.25 minutes on their smartphone each time they pick it up. if noelles distribution of smartphone use is approximately normal with a mean of 1.25 minutes and a 97th percentile of 2 minutes, calculate its standard deviation. standard deviation = minutes (round to 2 decimal places.)

Explanation:

Step1: Find the z - score for the 97th percentile

Using a standard normal distribution table or calculator, the z - score \(z\) corresponding to the 97th percentile (\(P(X\leq x)=0.97\)) is approximately \(z = 1.88\).

Step2: Use the z - score formula

The z - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(x = 2\) (the value at the 97th percentile), \(\mu=1.25\) (the mean), and \(\sigma\) is the standard deviation.
Substitute the values into the formula: \(1.88=\frac{2 - 1.25}{\sigma}\).

Step3: Solve for \(\sigma\)

First, simplify the numerator: \(2−1.25 = 0.75\).
Then, from \(1.88=\frac{0.75}{\sigma}\), we can solve for \(\sigma\) by cross - multiplying: \(\sigma=\frac{0.75}{1.88}\).
Calculate \(\frac{0.75}{1.88}\approx0.40\).

Answer:

\(0.40\)