According to a social media blog, time spent on a certain social networking website has a mean of 19 minutes per visit. assume that time spent on the social networking site per visit is normally distributed and that the standard deviation is 3 minutes. complete parts (a) through (d) below.

a. if you select a random sample of 25 sessions, what is the probability that the sample mean is between 18.5 and 19.5 minutes? (round to three decimal places as needed.)

b. if you select a random sample of 25 sessions, what is the probability that the sample mean is between 18 and 19 minutes? (round to three decimal places as needed.)

c. if you select a random sample of 100 sessions, what is the probability that the sample mean is between 18.5 and 19.5 minutes? (round to three decimal places as needed.)

d. explain the difference in the results of (a) and (c).

a) 0.593; b) 0.453; c) 0.904; d) The sample size is larger which raises the probability.

Step-by-step explanation:

We find the z score for each of these problems.  Each z score is for the mean of a sample rather than an individual value; this means we use the formula

The mean for each of these questions, μ, is 19; the standard deviation, σ, for each is 3.

For part a,

We want P(18.5 < X < 19.5).  Our sample size, n, is 25.  We find the z score of each endpoint, find the area under the curve to the left of each, and subtract them to find the area between the two values:

z = (18.5-19)/(3÷√25) = -0.5/(3÷5) = -0.5/0.6 = -0.83

z = (19.5-19)/(3÷√25) = 0.5/(3÷5) = 0.5/0.6 = 0.83

The area under the curve to the left of z = -0.83 is 0.2033, and the area under the curve to the left of z = 0.83 is 0.7967; this makes the area between them

0.7967-0.2033 = 0.5934 ≈ 0.593

For part b,

We want P(18 < X < 19).  Our sample size is 25.  

z = (18-19)/(3÷√25) = -1/(3÷5) = -1/0.6 = -1.67

z = (19-19)/(3÷√25) = 0/(3÷5) = 0/0.6 = 0

The area under the curve to the left of z = -1.67 is 0.0475, and the area under the curve to the left of z = 0 is 0.5000; this makes the area between them

0.5000 - 0.0475 = 0.4525 ≈ 0.453

For part c,

We want (18.5 < X < 19.5).  Our sample size is 100.

z = (18.5-19)/(3÷√100) = -0.5/(3÷10) = -0.5/0.3 = -1.67

z = (19.5-19)/(3÷√100) = 0.5/(3÷10) = 0.5/0.3 = 1.67

The area under the curve to the left of z = -1.67 is 0.0475, and the area under the curve to the left of z = 1.67 is 0.9515; this makes the area between them

0.9515 - 0.0475 = 0.904

For part d,

The sample size in part c is 4 times larger than that of part a.  This increases the probability.