l 2017 Problem 2 points) A gom ksows that each member, on average, spends 70 min

Question

l 2017 Problem 2 points) A gom ksows that each member, on average, spends 70 minutes at the gym per week with a standard deviation of 20 inutes. Assume the amount of time spends at the gym is nonmally distributed a) What is the probability that a randomly selectod customer spends less than 65 minutes at the gym? b) Suppose the gym surveys a random sample of 49 members about the amount of time spend at the gym each week. What are the expected value and standard deviation (standard enor) of the sample mean of the time spent at the gym ) If 49 members are randomly selected, what is the probability that the average time spent at the gym exceeds 75 minutes? la / Ke Page 10 of lS

Explanation / Answer

5.
NORMAL DISTRIBUTION
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 ~ N(0,1)
mean ( u ) = 70
standard Deviation ( sd )= 20
a.
probability that a randomly selected customer spends less than 65 minutes at the gym
P(X < 65) = (65-70)/20
= -5/20= -0.25
= P ( Z <-0.25) From Standard Normal Table
= 0.4013

b.
NORMAL DISTRIBUTION
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 ( u ) = 70
standard Deviation ( sd )= 20/ Sqrt ( 49 ) =2.8571
sample size (n) = 49
expected value = 70
standard deviation = 2.8571

c.
probability that average time spends at the gym exceeds 75 minutes
P(X > 75) = (75-70)/20/ Sqrt ( 49 )
= 5/2.857= 1.75
= P ( Z >1.75) From Standard Normal Table
= 0.0401

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

---

---

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