Question

2) A softball hitter has a 0.260 batting average: Each at-bat she has a 260 probability of getting a hit. Today she will get 7 at-bats. WTP she gets: a) exactly 2 hits? Formula: Substitution: Result: Tolone b) at most 3 hits? Calculator command: Result: c) at least 3 hits? Calculator command: Result: d) between 2 and 5 hits inclusive? Calculator command: med do Result:

Answer 1

This is a binomial distribution question with
n = 7
p = 0.26
q = 1 - p = 0.74
where

P(X = x) = I-ur bd


a)P(X = 2), n = 7, p=0.26 X~Binomial(n = 7.p = 0.26, X = 2) P(X = 2) = (0.2620.747-2 P(X = 2) = 21.00.2620.745 = 21.00.06760.2219 P(X = 2) = 0.31501018034304 0.315

binomcdf(7, 0.26, 3)

b)P(X<3), n = 7, p = 0.26 P(X <3) = P(0<x<3) P(X <3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) P(X S3) = (6)0.26°0.747-0+ (1) 0.26' 0.747–1+()0.2620.747–2+ () 0.26*0.747-3 P(X<3) = 10.26°) (0.747)+7.0(0.262)(0.746 +21.0(0.262)(0.745) +35.0(0.263) (0.744) P(X <3) = 1(1.0) (0.1215)+7.0(0.26) (0.1642)+21.0(0.0676)(0.2219)+35.0(0.0176)(0.2999) P(X <3) = 0.1215 + 0.2989+0.315 + 0.1845 PX < 3) = 0.9198

1 - binomcdf(7, 0.26, 2)

c)P(x > 3), n = 7, p = 0.26 P(X > 3) = P(3 < X < 7) P(x > 3) = 1 - P(X = 0) - P(X = 1) - P(X = 2) P(x > 3) = 1 - (0.26°0.747-0 + (1) 0.26+0.747-1 + (7)0.26°0.747-2) P(x > 3) = 1 - (1(0.26°) (0.747) + 7.0(0.26)(0.746) + 21.0(0.262)(0.745)) P(x > 3) = 1 -(1(1.0) (0.1215) + 7.0(0.26) (0.1642) +21.0(0.0676)(0.2219)) P(X > 3) = 1 - (0.1215 + 0.2989+ 0.315) P(X > 3) = 0.2646
binomcdf(7, 0.26, 5) - binomcdf(7, 0.26,1)

d)P(2<x<5) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) P2 < X < 5) = 0.2620.747-2 + 0.260.747-3 + 0.260.747-4 + 14) (0.260.747-5 5 P(2 < X < 5) = 21.0(0.262)(0.74") + 35.0(0.263)(0.744) + 35.0(0.264)(0.74") + 21.0(0.26 (0.74") P(2 < X <5) = 21.0(0.0676) (0.2219)+35.0(0.0176) (0.2999)+35.0(0.0046) (0.4052)+ 21.0(0.0012) (0.5476) P(2 < X <5) = 0.315+ 0.1845 +0.0648 + 0.0137 P(2 < X <5) = 0.578

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