Question

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). .

Answer 1

Answer:

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

$z=\frac{X-\mu}{\sigma \div \sqrt{n}}$

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.