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3 years ago

7

We have 55 independent normal observations all with mean 100. The first 50 observations have variance 76.4, and the last five ha

ve variance 127. (a) Calculate the probability that the first observation X1 is between 98 and 103. (b) Calculate the probability that the average X¯ = 1 55 P55 i=1 Xi is between 98 and 103.

1 answer:

yuradex [85]3 years ago

3 0

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

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