Answer:
0.2241 ; 0.9437
Step-by-step explanation:
Number of independent normal observations = 55
Mean(m) = 100
Variance of first 50 = 76.4
Variance of last five = 127
Probability that first observation is between 98 and 103
Zscore = x - m / sqrt(v)
For x = 103
Zscore = (103 - 100) /sqrt(76.4) = 0.34
For x = 98
Zscore = (98 - 100) / sqrt(76.4) = - 0.23
P(Z < - 0.23) = 0.4090
P(Z < 0.34) = 0.6331
0.6331 - 0.4090 = 0.2241
B) 1/n²Σ[(X1.V1) + (X2. V2)]
1/55²[(50*76.4) + (5*127)]
1/55² [3820 + 635]
1/55² [4455]
4455/3025
= 1.4727
Hence, variance of entire sample = 1.4727
X = 98 and 103
Zscore = x - m / sqrt(v)
For x = 103
Zscore = (103 - 100) /sqrt(1.4727) = 2.47
For x = 98
Zscore = (98 - 100) / sqrt(1.4727) = - 1.65
P(Z < - 1.65) = 0.0495
P(Z < 2.47) = 0.9932
0.9932 - 0.0495 = 0.9437
