Question

5. A tennis player makes a successful first serve 51% of the time. Assume that each serve is independent of the others. If she serves 9 times, what is the probability that she gets at most 3 successful first serves in?

6. The overhead reach distances of adult females are normally distributed with a mean of 202.5 cm and a standard deviation of 8.3 cm. Find the probability that the mean for 25 randomly selected distances is greater than 200.30 cm.

Answer 1

5)

Solution

Given that ,

p = 0.51

1 - p = 0.49

n = 9

Using binomial probability formula ,

P(X = x) = ((n! / x! (n - x)!) * p^x * (1 - p)^{n - x}

P(X 3 ) = 1 - P(X < 3 )

= 1 - ( P(X = 0) + P( X = 1 ) + P( X = 2 )

= 1 - ( ((9! / 0! (9-0)!) * 0.51⁰ * (0.49)^9-0 + ((9! / 1! (9-1)!) * 0.51¹ * (0.49)^9-1+ ((9! / 2! (9-2)!) * 0.51² * (0.49)^9-2)

= 1 - ( ((9! / 0! (9)!) * 0.51⁰ * (0.49)⁹ + ((9! / 1! (8)!) * 0.51¹ * (0.49)⁸ + ((9! / 2! (7)!) * 0.51² * (0.49)⁷)

= 1 - ( 0.0016 + 0.0153 + 0.0635)

= 1 - 0.0804

= 0.9196

Probability = 0.9196

6)

Given that ,

mean = = 202.5

standard deviation = = 8.3

n = 25

= = 202.5 and

= / n = 8.3 / 25 = 1.66

P( >200.30) = 1 - P( < 200.30)

= 1 - P(( - ) / < (200.30 - 202.5) / 1.66)

= 1 - P(z < -1.33)

= 1 - 0.0918

= 0.9082

Probability = 0.9082