Question

In a study on the physical activity of young adults, pediatric researchers measured overall physical activity as the total number of registered movements (counts) over a period of time and then computed the number of counts per minute (cpm) for each subject (International Journal of Obesity, Jan. 2007). The study revealed that the overall physical activity of obese young adults has a mean of m = 320 cpm and a standard deviation of s = 100 cpm. (In comparison, the mean for young adults of normal weight is 540 cpm.) In a random sample of n = 100 obese young adults, consider the sample mean counts per minute, x.

a. Describe the sampling distribution of x.

b. What is the probability that the mean overall physical activity level of the sample is between 300 and 310 cpm?

c. What is the probability that the mean overall physical activity level of the sample is greater than 360 cpm?

Answer 1

Solution:

Part a

Here, we have to describe the sampling distribution of X, the sample mean counts per minute.

We are given

Mean = µ = 320

Standard deviation = σ = 100

Sample size = n = 100

We know that best unbiased estimate of the mean of sampling distribution is population mean.

Mean of the sampling distribution = µ = 320

Standard deviation of the sampling distribution = σ/sqrt(n) = 100/sqrt(100) = 100/10 = 10

So, sampling distribution will follow an approximate normal distribution with mean 320 cpm with standard deviation of 10 cpm.

Part b

Here, we have to find P(300<Xbar<310)

P(300<Xbar<310) = P(Xbar<310) – P(Xbar<300)

First we have to find P(Xbar<310)

Z = (Xbar - µ)/[σ/sqrt(n)]

Z = (310 – 320)/[100/sqrt(10)]

Z = -10/10

Z = -1

P(Z<-1) = P(Xbar<310) = 0.158655

(by using z-table)

Now, we have to find P(Xbar<300)

Z = (300 – 320)/[100/sqrt(100)]

Z = -20/10

Z = -2

P(Z<-2) = P(Xbar<300) = 0.02275

(by using z-table)

P(300<Xbar<310) = P(Xbar<310) – P(Xbar<300)

P(300<Xbar<310) = 0.158655 – 0.02275

P(300<Xbar<310) = 0.135905

Required probability = 0.135905

Part c

Here, we have to find P(Xbar>360)

P(Xbar>360) = 1 – P(Xbar<360)

Z = (Xbar - µ)/[σ/sqrt(n)]

Z = (360 – 320)/[100/sqrt(10)]

Z = 40/10

Z = 4

P(Z<4) = P(Xbar<360) = 0.999968

(by using z-table)

P(Xbar>360) = 1 – P(Xbar<360)

P(Xbar>360) = 1 – 0.999968

P(Xbar>360) = 0.00003167

Required probability = = 0.00003167