Question

The scores of students on the SAT college entrance examinations at a certain high school had a normal distribution with mean μ=553.4μ=553.4 and standard deviation σ=29.3σ=29.3.

(a)What is the probability that a single student randomly chosen from all those taking the test scores 560 or higher?

For parts (b) through (d), consider a simple random sample (SRS) of 35 students who took the test.

(b)What are the mean and standard deviation of the sample mean score x¯x¯, of 35 students?
The mean of the sampling distribution for x¯x¯ is:

(c) What z-score corresponds to the mean score x¯x¯ of 560?

(d)What is the probability that the mean score x¯x¯ of these students is 560 or higher?

Answer 1

Part a)

X ~ N ( µ = 553.4 , σ = 29.3 )
P ( X >= 560 ) = 1 - P ( X < 560 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 560 - 553.4 ) / 29.3
Z = 0.2253
P ( ( X - µ ) / σ ) > ( 560 - 553.4 ) / 29.3 )
P ( Z > 0.2253 )
P ( X >= 560 ) = 1 - P ( Z < 0.2253 )
P ( X >= 560 ) = 1 - 0.5891
P ( X >= 560 ) = 0.4109

Part b)

Mean of sampling distribution

µ_X̅ = µ = 553.4

Standard deviation of sampling distribution
σ_X̅ = σ / √ (n) = 29.3/√35 = 4.9526

Part c)

X ~ N ( µ = 553.4 , σ = 29.3 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 560 - 553.4 ) / ( 29.3 / √ ( 35 ) )
Z = 1.33

Part d)

X ~ N ( µ = 553.4 , σ = 29.3 )
P ( X >= 560 ) = 1 - P ( X < 560 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 560 - 553.4 ) / ( 29.3 / √ ( 35 ) )
Z = 1.3326
P ( ( X - µ ) / ( σ / √ (n)) > ( 560 - 553.4 ) / ( 29.3 / √(35) )
P ( Z > 1.33 )
P ( X̅ >= 560 ) = 1 - P ( Z < 1.33 )
P ( X̅ >= 560 ) = 1 - 0.9087
P ( X̅ >= 560 ) = 0.0913