A study was conducted on the

minutes people spend browsing

websites. The time spent is normally

distributed with a mean of 4.5

minutes and a standard deviation of

1.2 About what percentage of

people spent between 4 and 6

minutes browsing the sites? *

Step-by-step explanation:

Let x be the random variable representing the minutes people spend browsing websites. Since the population mean and population standard deviation are known and it is a normal distribution, we would apply the formula,

z = (x - µ)/σ

Where

x = sample mean

µ = population mean

σ = standard deviation

From the information given,

µ = 4.5 minutes

σ = 1.2 minutes

The probability that people spent between 4 and 6 minutes browsing the sites is expressed as

P(4 ≤ x ≤ 6)

For x = 4,

z = (4 - 4.5)/1.2 = - 0.42

Looking at the normal distribution table, the probability corresponding to the z score is 0.3372

For x = 6

z = (6 - 4.5)/1.2 = 1.25

Looking at the normal distribution table, the probability corresponding to the z score is 0.8944

Therefore,

P(4 ≤ x ≤ 6) = 0.8944 - 0.3372 = 0.5572

The percentage of people spending between 4 and 6 minutes browsing the sites

0.5572 × 100 = 55.72%

answer: you need the equation in the question so i can answer it

step-by-step explanation: