
Jon S. answered 04/01/21
Patient and Knowledgeable Math and English Tutor
P(85 < xbar < 123)
transform above range to N(0,1) by computing z-scores:
P( (85-105)/(20/sqrt(19)) < z < (123-105)/(20/sqrt(19)) = P(-4.36 < z < 3.92) ~ 1
Kim B.
asked 04/01/21A typical adult has an average IQ score of 105 with a standard deviation of 20. If 19 randomly selected adults are given an IQ test, what is the probability that the sample mean scores will be between 85 and 123 points? (Round your answer to five decimal places.)
Jon S. answered 04/01/21
Patient and Knowledgeable Math and English Tutor
P(85 < xbar < 123)
transform above range to N(0,1) by computing z-scores:
P( (85-105)/(20/sqrt(19)) < z < (123-105)/(20/sqrt(19)) = P(-4.36 < z < 3.92) ~ 1
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