Question

One college class had a total of 70 students. The average score for the class on the last exam was 84.2 with a standard deviation of 4.8. A random sample of 31 students was selected. a. Calculate the standard error of the mean. b. What is the probability that the sample mean will be less than 86​? c. What is the probability that the sample mean will be more than 85​? d. What is the probability that the sample mean will be between 83.5 and 85.5​? a. The standard error of the mean is nothing. ​(Round to two decimal places as​ needed.) b. The probability that the sample mean will be less than 86 is nothing. ​(Round to four decimal places as​ needed.) c. The probability that the sample mean will be more than 85 is nothing. ​(Round to four decimal places as​ needed.) d. The probability that the sample mean will be between 83.5 and 85.5 is nothing. ​(Round to four decimal places as​ needed.)

Answer 1

Solution :

Given that ,

mean = = 84.2

standard deviation = = 4.8

n = 31

a) =    = 84.2

= / n = 4.8 / 31 = 0.86

b) P( < 86 ) = P(( - ) / < ( 86 - 84.2) / 0.86 )

= P(z < 2.09 )

Using z table

= 0.9817

c) P( > 85 ) = 1 - P( < 85 )

= 1 - P[( - ) / < ( 85 - 84.2) / 0.86 ]

= 1 - P(z < 0.93 )

Using z table

= 1 - 0.8238

= 0.1762

d) P( 83.5 < < 85.5 )

= P[( 83.5 - 84.2 ) / 0.86 < ( - ) / < ( 85.5 - 84.2 ) / 0.86 )]

= P( -0.81 < Z < 1.51 )

= P(Z < 1.51 ) - P(Z < -0.81 )

Using z table

= 0.9345 - 0.2090

=0.7255