In: Math
1. Kelly and Veronica are two teachers in a math class who attend class independently of one another. For Friday classes, there is a .70 probability that kelly will come to class, while there is a .40 probability that Veronica will come to class. For a Friday class, what is the probability neither Kelly nor Veronica will be there?
2. The weights of newborn baby twin girls have an approximately normal distribution with a mean of 8.0 pounds and a standard deviation of 1.5 pounds. A doctor tells the family that one of the baby twin girl has a weight at the 30th percentile. Which of the following is closest to this baby's weight? (show work please)
A, 7.2
B 8.5
C 7.7
D 8.9
1.
Let, K be the event that Kelly will come to class
Let, V be the event that Veronica will come to class
Probability that Kelly will come to class = P(K) = 0.7
Probability that Kelly will not come to class = P(K) = 1 - P(K) = 1 - 0.7 = 0.3
Probability that Veronica will come to class = P(V) = 0.4
Probability that Veronica will not come to class = P(V) = 1 - P(V) = 1 - 0.4 = 0.6
Since, 1. Kelly and Veronica are two teachers in a math class attend classes independently of one another
Probability neither Kelly nor Veronica will be there = Probability that Kelly will not come to class*Probability that Veronica will not come to class = P(K)*P(V) = 0.6*0.3 = 0.18
Probability neither Kelly nor Veronica will be there = 0.18
2. Let X be the weights of newborn baby twin girls have an approximately normal distribution
mean of X , E(X) = 8.0 pounds
Standard deviation = S(X) =1.5 pounds.
30th percentile mean 30% girls weigh lesser than that of her.
Let, 30th percentile be p_30
P[ X < p_30 ] = 30% = 0.3
X ~ N(8,2.25)
P[ X < p_30 ] = 0.3
P[ ( X - E(X) )/S(X) < ( p_30 - E(X) )/S(X) ] = 0.3
P[ ( X - 8 )/1.5 < ( p_30 - 8 )/1.5 ] = 0.3
P[ Z < ( p_30 - 8 )/1.5 ] = 0.3
Also, we know that
P[ Z < -0.52 ] = 0.3
Comparing, ( p_30 - 8 )/1.5 = -0.52
p_30 - 8 = 1.5*(-0.52)
p_30 - 8 = -0.78
p_30 = 8 - 0.78
p_30 = 7.22
Correct answer is a) = 7.2 pounds