Question

The composition scores of individuals on the ACT college entrance examination in 2019 have mean 20.8 and standard deviation 5.8. Use this information, answer the following questions: (Round z-score values to 2 decimal places) (a) Now suppose that the composition scores are normally distributed, what is the probability that a single student randomly chosen from all those taking the test scores higher than 22? (6 points) (b) (Ignore part a) Now suppose that we take a simple random sample of 100 students who took the test, answer parts (i) to (iii): (i) Why are we allowed to use the normal approximation for the sampling distribution of ?̅? (4 points) (ii) Find the mean and standard deviation of the sampling distribution of ?̅. (4 points) (iii) What is the probability that the mean score of these 100 students is higher than 22? (6 points)

Answer 1

Solution :

Given that ,

mean = = 20.8

standard deviation = = 5.8

a) P(x > 22) = 1 - p( x< 22)

=1- p P[(x - ) / < (22 - 20.8) / 5.8]

=1- P(z < 0.21)

Using z table,

= 1 - 0.5832

= 0.4168

b) n = 100

(i) The sampling distribution of is approximately normal, because sample size greater than 30

(ii) = = 20.8

= / n = 5.8 / 100 = 0.58

(iii) P( > 22) = 1 - P( < 22)

= 1 - P[( - ) / < (22 - 20.8) / 0.58 ]

= 1 - P(z < 2.07)

Using z table,    

= 1 - 0.9808

= 0.0192

Answer 2

Solution :

Given that ,

mean = = 20.8

standard deviation = = 5.8

a) P(x > 22) = 1 - p( x< 22)

=1- p P[(x - ) / < (22 - 20.8) / 5.8]

=1- P(z < 0.21)

Using z table,

= 1 - 0.5832

= 0.4168

b) n = 100

(i) The sampling distribution of is approximately normal, because sample size greater than 30

(ii) = = 20.8

= / n = 5.8 / 100 = 0.58

(iii) P( > 22) = 1 - P( < 22)

= 1 - P[( - ) / < (22 - 20.8) / 0.58 ]

= 1 - P(z < 2.07)

Using z table,    

= 1 - 0.9808

= 0.0192