If u = 100, sigma = 12, and n = 16, then use the t distribution...

If u = 100, sigma = 12, and n = 16, then use the t distribution to evaluate the probability ? exceeds 103?

Express your answer as a percentage with 3 decimals

Solutions

Expert Solution

Solution :

mean = = 100

standard deviation = = 12

n = 16

= = 100

= / n = 12 / 16 = 3

P( > 103) = 1 - P( < 103)

= 1 - P[( - ) / < (103 - 100) / 3]

= 1 - P(z < 1)

Using z table,

= 1 - 0.8413

= 0.1587

The percentage is 15.87%

orchestra answered 1 year ago

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