Question

suppose the population of teacher incomes in SD is strongly skewed to the right with a mean of $50,000 and a standard deviation of $10,000. If you use a sample size of 64, what is the probability that the sample mean will exceed $52,750? [HINT -use the Z table in conjunction with the values for the sampli ng distribution model]. (A)0.0139 (B)0.3897 (C)0.6103 (D)0.9861

Answer 1

Solution :

Given that ,

mean = = 50000

standard deviation = = 10000

n = 64

= 50000 and

= / n = 10000 / 64 = 10000 / 8 = 1250

P( > 52750) = 1 - P( < 52750)

= 1 - P(( - ) / < (52750 - 50000) / 1250)

= 1 - P(z < 2.2)

= 1 - 0.9861

= 0.0139

P( > 52750) = 0.0139

Probability = 0.0139