Question

The time for visitor to read health instructions on a Web site is approximately normally distributed with a mean of 10 minutes and a standard deviation of 2 minutes. Suppose 64 visitors independently view the site. Determine the following:

1. Expected value and variance of the mean time of the visitors

2. Probability that the mean time of the visitors is within 15 seconds of 10 minutes

3. Value exceeded by the mean time of the visitors with probability 0.01.

Answer 1

Answer:

Given,

Mean = 10

Standard deviation = 2

Sample n = 64

a)

Mean = ux = 10

Sandard deviation = s/sqrt(n)

substitute values

= 2/sqrt(64)

= 0.25

Variance = (standard deviation)^2

= 0.25^2

= 0.0625

b)

Here 10 +/- 10/60

= 10 +/- 0.25

= (9.75 , 10.25)

Required probability = P(9.75 < X < 10.25)

= P((9.75 - 10)/0.25 <(x-u)/s < (10.25 - 10)/0.25)

= P(-1 < z < 1)

= P(z < 1) - P(z < -1)

= 0.8413447 - 0.1586553 [since from z table]

= 0.6827

c)

Given,

P(X < x) = 0.01

Here from the standard normal table

z = - 2.33

So, x = u + z*s

substitute values

= 10 - 2.33*0.25

= 10 - 0.5825

= 9.4175