a normal population has a mean of 80 and standard deviation of 3. you select a s

Question

a normal population has a mean of 80 and standard deviation of 3. you select a sample of 44. compute the probability the sample mean a. Less than 79
b. Between 79 and 81
c. Between 81 and 82
d. Greater than 82 a normal population has a mean of 80 and standard deviation of 3. you select a sample of 44. compute the probability the sample mean a. Less than 79
b. Between 79 and 81
c. Between 81 and 82
d. Greater than 82 a normal population has a mean of 80 and standard deviation of 3. you select a sample of 44. compute the probability the sample mean a. Less than 79
b. Between 79 and 81
c. Between 81 and 82
d. Greater than 82

Explanation / Answer

Given data:

Mean, m = 80

SD, S = 3

Sample size, n = 44

Standard error, SE = S/n0.5 = 3/440.5 = 0.452

(a)

At value of 79, z-score = (79-m)/SE = (79-80)/0.452 = -2.21

Using the cumulative z-table the probability value is: p = 0.013

(b)

At value of 79, z-score = (79-80)/0.452 = -2.21

Using the cumulative z-table the probability value is: p1 = 0.013

At value of 81, z-score = (81-80)/0.452 = 2.21

Using the cumulative z-table the probability value is: p2 = 0.986

So, required probability = p2-p1 = 0.986-0.013 = 0.973

(c)

At value of 81, z-score = (81-80)/0.452 = 2.21

Using the cumulative z-table the probability value is: p1 = 0.986

At value of 82, z-score = (82-80)/0.452 = 4.42

Using the cumulative z-table the probability value is: p2 = 0.999

So, required probability = p2-p1 = 0.999-0.986 = 0.013

(d)

At value of 82, z-score = (82-80)/0.452 = 4.42

Using the cumulative z-table the probability value is: p = 0.999

So, required probability = 1-p = 1-0.999 = 0.001

Hope this helps !

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