1 When a group of 5 single births are recorded what is the p

1. When a group of 5 single births are recorded, what is the probability that at least 4 are boys? Assume that boys and girls are equally likely. (Hint: You need a probability distribution)

a) 80% b) 15.62% c) 18.75% d) 81.25% e) 74.38%

2. One of these is not a requirement of a binomial distribution:

a) A fixed number of trials. b) All trials are independent.

c) Several possible outcomes. d) The probabilities remain the same through each trial.

3. Given the following probability Distribution, find the mean

X P(X)

0 0.422

1 0.422

2 0.141

3 0.016

a) 0.550 b) 0.752 c) 0.348 d) 0.702 e) 0.657

4. Find the standard deviation using the table from question 3

a) 0.7513 b) 1.006 c) 0.5644 d) 0.4356 e) 0.8743

Solution

1)

Note that P(at least x) = 1 - P(at most x - 1).          
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    5      
p = the probability of a success =    0.5      
x = our critical value of successes =    4      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   3   ) =    0.8125
          
Thus, the probability of at least   4   successes is  
          
P(at least   4   ) =    0.1875 = 18.75% [ANSWER, OPTION C]

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2.

c) Several possible outcomes [actually, there should be just 2]

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3.

Consider the table:

Thus,          
          
E(x) = Sum(xP(x)) =    0.752 [ANSWER, OPTION B]

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4.

Here,

E(X) = 0.752

E(x^2) = 1.13

Thus,

Var(x) = E(x^2) - E(x)^2 =    0.564496 [ANSWER, OPTION C]