HW9 Solutions

6.16

1.  = 10,  = 0.6

2.  = 100,  = 5

3.  = 20,  = 8

4.  = 10,  = 20

6.18

No. Only when the sample is randomly selected and either the parent population is normally distributed or the sample size is large enough for the Central Limit Theorem to hold.

6.20

1. P( < 16) =
   P[z < (16-20)/(16/8)] =
   P(z < -2) = 0.0228
    

2. P( > 23) =
   P[z > (23-20)/(16/8)] =
   P(z > 1.5) = 0.0668
    

3. P( > 25) =
   P[z > (25-20)/(16/8)] =
   P(z > 2.5) = 0.0062
   

4. P(16 <  < 22) =
   P[(16-20)/(16/8) < z < (22-20)/(16/8)] =
   P(-2 < z < 1) =
   1 - 0.0228 - 0.1587 = 0.8185
    

5. P( < 14) =
   P[z < (14-20)/(16/8)] =
   P(z < -3) = 0.0013

6.24

1.  = 3.5,  = 0.05
    

2. P(3.40 <  < 3.60) =
   P[(3.4-3.5)/0.05 < z < (3.6-3.5)/0.05] =
   P(-2 < z < 2) =
   1 - 0.0228 - 0.0228 = 0.9544
    

3. P( > 3.62) =
   P[(z > (3.62-3.5)/0.05] =
   P(z > 2.4) =
   0.0082
    

4. The standard error would get smaller, meaning the variance of the sampling distribution would decrease. The mean of the distribution wouldn't change. Problem b) would be 0.9954 (the probability would increase0 and problem c) would change to 0 (probability would get smaller) 

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