Question

Answer 1

ANSWER :-

Given,
Sample Size, n = 89
Sample Mean, Xbar = 15.83
Sample Standard Deviation, S = 1.74

1)
P(14.09 < X < 17.57) = P[ (14.09-Xbar)/S < (X-Xbar)/S < (17.57-Xbar)/S ]
= P[ (14.09-15.83)/1.74 < Z < (17.57-15.83)/1.74 ]
= P(-1 < Z < 1)
= P(Z < 1) - P(Z < -1)
= 0.841345 - 0.158655
= 0.6827
Percentage of the fifth grade classes contain “student-to-faculty ratios” from 14.09 to 17.57 is 68.27%

2)
P(-z < Z < z) = 0.95
=> P(Z < z) - P(Z < -z) = 0.95
=> P(Z < z) - [ 1 - P(Z < z) ] = 0.95
=> P(Z < z) - 1 + P(Z < z) = 0.95
=> 2*P(Z < z) = 1+0.95
=> P(Z < z) = 1.95/2
=> P(Z < z) = 0.975
From Z table
P(Z < 1.96) = 0.975

=> z = 1.96
=> (X-Xbar)/S = 1.96
=> (X-15.83)/1.74 = 1.96
=> X = (1.96*1.74)+15.83
=> X = 19.2404

-z = -1.96
=> (X-Xbar)/S = -1.96
=> X = (-1.96*1.74)+15.83
=> X = 12.4196

95% of the fifth grade classes contain “student-to-faculty ratios” from 12.4196 to 19.2404

3)
P(10.61 < X < 21.05) = P[ (10.61-Xbar)/S < (X-Xbar)/S < (21.05-Xbar)/S ]
= P[ (10.61-15.83)/1.74 < Z < (21.05-15.83)/1.74 ]
= P(-3 < Z < 3)
= P(Z < 3) - P(Z < -3)
= 0.99865 - 0.00135
= 0.9973

P*n = 0.9973*81 = 80.7813
80.78 fifth grade classes contain “student-to-faculty ratios” from 10.61 to 21.05 classes.