This question is incomplete
Complete Question
A normal population has a mean of 61 and a standard deviation of 13. You select a random sample of 16. Compute the probability that the sample mean is: (Round z values to 2 decimal places and final answers to 4 decimal places.)
(a) Greater than 64
(b) Less than 57
Answer:
(a) Greater than 64 = 0.1788
(b) Less than 57 = 0.1094
Step-by-step explanation:
To solve the above questions we would be using the z score formula
The formula for calculating a z-score :
z = (x - μ)/σ,
where x is the raw score
μ is the population mean = 61
σ is the population standard deviation = 13
(a) Greater than 64
z = (x - μ)/σ,
where x is 64
μ is the 61
σ is the 13
In the above question, we are given the number of samples = 16
Sample standard deviation = popular standard deviation/ √16
= 13/√16
z = 64 - 61 ÷ 13/√16
z = 3/3.25
z = 0.92308
Approximately, z values to 2 decimal places ≈ 0.92
Using the z score table of normal distribution to find the Probability (P) value of z score of 0.92
P(z = 0.92) = 0.82121
P(x>64) = 1 - P(z = 0.92)
= 1 - 0.82121
= 0.17879
Approximately , Probability value to 4 decimal places = 0.1788
(b) Less than 57
z = (x - μ)/σ,
where x is 57
μ is the 61
σ is the 13
In the above question, we are given the number of samples = 16
Sample standard deviation = popular standard deviation/ √16
= 13/√16
z = 57 - 61 ÷ 13/√16
z = -4/3.25
z = -1.23077
Approximately, z values to 2 decimal places ≈ -1.23
Using the z score table of normal distribution to find the Probability (P) value of z score of -1.23
P(z = -1.23) = P(x<Z) = 0.10935
Approximately , Probability value to 4 decimal places = 0.1094
