1 answer

A simple random sample of size n 49 is obtained from a population that is skewed...

Question:

A simple random sample of size n 49 is obtained from a population that is skewed right with u= 74 and o 28. (a) Describe the

A simple random sample of size n 49 is obtained from a population that is skewed right with u= 74 and o 28. (a) Describe the sampling distribution of x. (b) What is P (x>81.8)? (c) What is P Xs65.6)? (d) What is P (70.4 <x<81.2)?

Answers

Part a)

sampling distribution of X̅

µ = µ = 74
σ = σ / √ (n) = 28/√49 = 4

Part b)

X ~ N ( µ = 74 , σ = 28 )
P ( X̅ > 81.8 ) = 1 - P ( X̅ < 81.8 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 81.8 - 74 ) / ( 28 / √ ( 49 ) )
Z = 1.95
P ( ( X - µ ) / ( σ / √ (n)) > ( 81.8 - 74 ) / ( 28 / √(49) )
P ( Z > 1.95 )
P ( X̅ > 81.8 ) = 1 - P ( Z < 1.95 )
P ( X̅ > 81.8 ) = 1 - 0.9744
P ( X̅ > 81.8 ) = 0.0256


Part c)
X ~ N ( µ = 74 , σ = 28 )
P ( X̅ <= 65.6 )
Standardizing the value
Z = ( X - µ ) / (σ/√(n)
Z = ( 65.6 - 74 ) / ( 28 / √49 )
Z = -2.1
P ( ( X - µ ) / ( σ/√(n)) < ( 65.6 - 74 ) / ( 28 / √(49) )
P ( X <= 65.6 ) = P ( Z < -2.1 )
P ( X̅ <= 65.6 ) = 0.0179


Part d)
X ~ N ( µ = 74 , σ = 28 )
P ( 70.4 < X̅ < 81.2 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 70.4 - 74 ) / ( 28 / √(49))
Z = -0.9
Z = ( 81.2 - 74 ) / ( 28 / √(49))
Z = 1.8
P ( -0.9 < Z < 1.8 )
P ( 70.4 < X̅ < 81.2 ) = P ( Z < 1.8 ) - P ( Z < -0.9 )
P ( 70.4 < X̅ < 81.2 ) = 0.9641 - 0.1841
P ( 70.4 < X̅ < 81.2 ) = 0.78

.

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