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