Question

The composite scores of individual students on the ACT college entrance examination in 2009 followed a normal distribution with mean 21.1 and standard deviation 5.1.

1. What is the probability that a single student randomly chosen from all those taking the test scores 23 or higher?
   P( X >__?__ ) = P(z >__?__ ) = ____?__
2. 
   
3. What is the probability that a simple random sample of 50 students chosen from all those taking the test has an average score of 23 or higher?
   P( X̄ >_?__ ) = P(z >___?__ ) = ___?____

Answer 1

Solution :

Given that ,

mean = = 21.1

standard deviation = = 5.1

a) P(x ≥ 23 ) = 1 - P(x  ≤ 23)

= 1 - P[(x - ) / ≤  (23-21.1) /5.1 ]

= 1 -  P(z ≤ 0.37)  

= 1 - 0.6443 = 0.3557

Probability= 0.3557

b)

n = 50

=   = 21.1

= / n = 5.1/ 50 = 0.7212

P( ≥ 23) = 1 - P( ≤ 23 )

= 1 - P[( - ) / ≤ (23 - 21.1) / 0.7212 ]

= 1 - P(z ≤ 2.63)

= 1 - 0.9957 = 0.0043

Probability = 0.0043