Question 1021385: 1. Let discrete random variable X: # of boys for a family with 7 children.
b. Find the probability of having exactly 4 boys.
c. Find the whole probability distribution.
d. Find the probability of having at most 3 boys.
e. Find the probability of having at least 5 boys.
2. Find the mean and the standard deviation of the experiment on problem 1. Would it be unusual to have only 2 boys?
3. A basketball player averages 76% from the free-throw line. If this player takes 250 shots from the free-throw line:
a. What are the chances of him making 190 shots of those 250 shots.
b. What are the chances of him making at least 200 shots?
c. What are the chances of him making at most 220 shots?
4. If you roll two dice 50 times, what are the chances of you rolling doubles (both numbers on the dice are the same) 15 times?
Answer by mathmate(429)   (Show Source):
You can put this solution on YOUR website!
Question:
1. Let discrete random variable X: # of boys for a family with 7 children.
b. Find the probability of having exactly 4 boys.
c. Find the whole probability distribution.
d. Find the probability of having at most 3 boys.
e. Find the probability of having at least 5 boys.
2. Find the mean and the standard deviation of the experiment on problem 1. Would it be unusual to have only 2 boys?
3. A basketball player averages 76% from the free-throw line. If this player takes 250 shots from the free-throw line:
a. What are the chances of him making 190 shots of those 250 shots.
b. What are the chances of him making at least 200 shots?
c. What are the chances of him making at most 220 shots?
4. If you roll two dice 50 times, what are the chances of you rolling doubles (both numbers on the dice are the same) 15 times?
Solution:
1. This is a case of the binomial expansion of (B+G)^7 with the following coefficients for B^i: C={1,7,21,35,35,21,7,1} the sum of which is 2^7=128.
i ranges from 0 to 7.
(a) The corresponding distribution is therefore C/128, or
p(X)={1/128,7/128,21/128,35/128,35/128,21/128,7/128,1/128}
(b)from P(X) above, P(X=4)=35/128
(c) see (a) above
(d) P(X<=3)=(1+7+21+34)/128=63/128
(e) P(X>=5)=(21+7+1)/128=29/128
2. For a binomial distribution, mean=np, variance = npq, and P(X;n;p)=P(X=k)=
n=number of trials (7)
p=probability of success = 0.5 (assumed)
q=probability of failure = 1-p = 0.5
C(n,k)=n!/(k!(n-k)!) is the number of combinations of k objects out of n.
So
mean = μ = 7*0.5=0.35
standard deviation = sqrt(variance) = sqrt(7*0.5*0.5)=1.323
3. Again, binomial distribution, with p=0.76, n=250.
note: use a calculator with binomial distribution function, or an application such as R. Statements in [] appliy to R commands.
(a) exactly 190 shots
P(190;250;0.76)=0.05899 [dbinom(190,250,0.76)]
(b) at least 200 shots
P(>=200;250;0.76)=0.07768 [ pbinom(199,250,0.76,lower.tail=FALSE) ]
note: the upper-tail in R is defined as P(X>199) which equals P(X>=200)
(c) at most 220 shots
P(<=220;250;0.76)=0.999999411 [pbinom(220,250,0.76)]
Alternate solution: if an application is not available, the probabilities can be approximated by the normal distribution, with μ=np=190, σ=sqrt(npq)=6.753. The continuity correction should be applied.
4. Probability of rolling a double in a single throw is p=6/36=1/6.
If the dies are rolled 50 times, n=50, p=1/6, q=1-1/6=5/6.
P(15;50;1/6)=C(50,15)*(1/6)^15*(5/6)^(50-15)
=0.008105 [dbinom(15,50,1/6)]
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