2 According to research data based on weather age and course

2. According to research data based on weather, age, and course, the population for time that it takes a person to run a 5K is normally distributed with a mean of 35.0 minutes and a standard deviation of 3.16 minutes.

a) What is the mean and standard deviation of the sampling distribution of the sample mean for samples of size 16?

b) If a sample of 16 individuals is chosen at random, what is the probability that the sampling error between the sample mean ( x ) and the population mean ( ) is less than one minute?

Solution

a)
Sample Mean = Population Mean ( u ) = 35
Standard Deviation ( sd )= (sd/Sqrt(n)) = 3.16/ Sqrt ( 16 ) = 0.79
Number ( n ) = 16
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)
b)
Sampling Error= sd/ Sqrt(n)
Where,
sd = Standard Deviation
n = Sample Size

Standard deviation( sd )= 0.79
Sampling Error that less than 1 mins = P(35-1 < X < 35+1)= P(34 < X < 36)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 34) = (34-35)/3.16/ Sqrt ( 16 )
= -1/0.79
= -1.2658
= P ( Z <-1.2658) From Standard Normal Table
= 0.10279
P(X < 36) = (36-35)/3.16/ Sqrt ( 16 )
= 1/0.79 = 1.2658
= P ( Z <1.2658) From Standard Normal Table
= 0.89721
P(34 < X < 36) = 0.89721-0.10279 = 0.7944