12% D mathxl.com DS 2333 Fall 2017 Chris Mazza 11/13/17 4:03 Homework: Chapter 7

Question

12% D mathxl.com DS 2333 Fall 2017 Chris Mazza 11/13/17 4:03 Homework: Chapter 7: Sec. 1, 2, & 3 Score: 0 of 1 pt 7.2.7-T 3015(2 complete) HW Score: 51.43%, 3.6 Question Help Time spent using e-mail per session is normally distributed, with-11 minutes and -2 minutes. Assume that the time spent per session is normally distributed. Complete parts (a) through (d). a. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes? 0.383 (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 10.5 and 11 minutes? (Round to three decimal places as needed.) Enter your answer in the answer box and then click Check Answer. Clear All remaining

Explanation / Answer

the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean of the sampling distribution ( x ) = 11
standard Deviation ( sd )= 2/ Sqrt ( 25 ) =0.4
sample size (n) = 25
a)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 10.8) = (10.8-11)/2/ Sqrt ( 25 )
= -0.2/0.4
= -0.5
= P ( Z <-0.5) From Standard Normal Table
= 0.30854
P(X < 11.2) = (11.2-11)/2/ Sqrt ( 25 )
= 0.2/0.4 = 0.5
= P ( Z <0.5) From Standard Normal Table
= 0.69146
P(10.8 < X < 11.2) = 0.69146-0.30854 = 0.383
b)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 10.5) = (10.5-11)/2/ Sqrt ( 25 )
= -0.5/0.4
= -1.25
= P ( Z <-1.25) From Standard Normal Table
= 0.10565
P(X < 11) = (11-11)/2/ Sqrt ( 25 )
= 0/0.4 = 0
= P ( Z <0) From Standard Normal Table
= 0.5
P(10.5 < X < 11) = 0.5-0.10565 = 0.394

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.