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If a baseball player's batting average is 0.340 or 34%, find the probability that the player...

If a baseball player's batting average is 0.340 or 34%, find the probability that the player will have a bad season and only score at most 60 hits in 200 times at bat?

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Answer #1

X ~ Binomial(n-200, p-0.34) Binomial can be approximated to normal with np 200* 0.34 -68 σーyMp(1-p)-V 200 (0.34) (1 0.34) 6.6

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Answer #2
Binomial Problem with n= 200 ; p = 0.34 ; x =60

By using normal approximation,
X~Normal(μ=n*p=200*0.34=68,σ=√n*p*(1-p)=√200*0.34*(1-0.34)=6.7)

So the probability is
P(X<60)
=P((X-μ)/σ < (60-68)/6.7)
=P(Z< -1.19)
=0.117 (check normal table)
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Answer #3
sample size, n= 200
proportion,p = 0.34

mean, μ=n*p=200*0.34=68,
standard deviation,σ=√n*p*(1-p)=√200*0.34*(1-0.34)=6.7

So the probability the player will have a bad season and only scoreat most 60 hits in 200 times at bat

P(X60)
=P()
=P( z -1.19)
=0.117                            (from z table)
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Answer #4
Step 1: Define terms.
p = population proportion =0.34
ph = sample proportion (or p hat) = 60/200 = 0.30
n = sample size = 200
Step 2: Find the standard error.
SE(ph) =standard error of p hat = √[ p * (1 – p) / n]
   = √(0.34 * 0.66 / 200) = 0.03350
Step 3: Find the z-value.
z = (ph – p) /SE(ph)
    = ( 0.30 – 0.34) / 0. 03350 = -1.19
Step 4: Find the probability.
P(ph < 0.30)= P(Z < -1.91) =0.1162
Hope this helps,
Mike
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