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: (

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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

(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

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