A certain tennis player makes a successful first serve 61% of the time. Suppose the tennis player serves 70 times in a match. What's the probability that she makes at least 56 first serves? The probability she makes at least 56 first serves is nothing.
Solution :
Given that,
p = 0.61
q = 1 - p =1-0.61=0.39
n = 70
Using binomial distribution,
= n * p = 70*0.61=42.7
= n * p * q = 70*0.61*0.39=4.0808
Using continuity correction
,P(x >56 ) = 1 - P(x <55.5 )
= 1 - P((x - ) / < (55.5-42.7) / 4.0808)
= 1 - P(z <3.14 )
Using z table
= 1-0.9992
probability= 0.0008
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