In: Statistics and Probability
One college class had a total of 80 students. The average score for the class on the last exam was 83.9 with a standard deviation of 5.8. A random sample of 32 students was selected. a. Calculate the standard error of the mean. b. What is the probability that the sample mean will be less than 85? c. What is the probability that the sample mean will be more than 84? d. What is the probability that the sample mean will be between 82.5 and 84.5? 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 85 is nothing. (Round to four decimal places as needed.) c. The probability that the sample mean will be more than 84 is nothing. (Round to four decimal places as needed.) d. The probability that the sample mean will be between 82.5 and 84.5 is nothing. (Round to four decimal places as needed.)
Solution :
Given that ,
mean = = 83.9
standard deviation = = 5.8
n = 32
= = 83.9
a) = / n = 5.8 / 32 = 1.03
b) P( < 85) = P(( - ) / < (85 - 83.9) / 1.03)
= P(z < 1.07)
Using z table
= 0.8577
c) P( > 84) = 1 - P( < 84)
= 1 - P[( - ) / < (84 - 83.9) / 1.03 ]
= 1 - P(z < 0.10)
Using z table,
= 1 - 0.5398
= 0.4602
d) P(82.5 < < 84.5)
= P[(82.5 - 83.9) /1.03 < ( - ) / < (84.5 - 83.9) / 1.03)]
= P(-1.36 < Z < 0.58)
= P(Z < 0.58) - P(Z < -1.36)
Using z table,
= 0.7190 - 0.0869
= 0.6321