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
Answer:
Quadrant 4
Step-by-step explanation:
Look at the picture. I highlighted quadrant 4. You can see that the graph did not pass the highlighted area.
16 sandwiches. Each one with two slices of bread and one slice of cheese.
If you mean 720/10^3, you do 10*10*10. This equals 1000. Now, do 720/1000. This is 0.72, or 72/100, which is your answer.