Batting averages. Suppose that a baseball player’s long-run batting average (number of hits per time at bat) is 300. Assuming that each time at bat yields a hit with a consisted probability,...

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Batting averages. Suppose that a baseball player’s long-run batting average (number of hits per time at bat) is 300. Assuming that each time at bat yields a hit with a consisted probability, independently of other times, what is the chance that the player’s average over the next 100 times at bat will be


a)       310 or better?                   b) .330 or better?            c)  .270 or worse?


d)      Suppose the player tends to have periods of good form and periods of bad form would different times at bat then be independent? Would that tend to increase or decreases the above chances?


e)      Suppose the player actually hits .330 over the 100 times at bat. Would you be convinced that his form had improved significantly? Or could the improvement just as well be due to chance?



Answered Same DayDec 26, 2021

Answer To: Batting averages. Suppose that a baseball player’s long-run batting average (number of hits per time...

Robert answered on Dec 26 2021
120 Votes
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in
Given: p = 0.3, q = 1− p = 0.7,n = 100
Taking normal ap
proximation to binomial, the sampling distribution of sample
proportion follows normal distribution.
Shape: Normal
p̂ ∼ N(µ,σ2), where
mean, µp̂ = p = 0.3
µp̂ = 0.3
variance, σ2 = p× (1− p)
n
= 0.3× 0.7100 = 0.0021
standard deviation, σp̂ =

σ2 =

0.3× 0.7
100 ≈ 0.045826
σp̂ = 0.045826
p̂ ∼ N(0.3, 0.0458262)
A P (p̂ ≥ 0.31)
Normal Distribution, µ = 0.3, σ = 0.045826,
Since this is continuous distribution, P (p̂ ≥ 0.31) = P (p̂ > 0.31)
we convert this to standard normal using z = p̂− µ
σ
z = 0.31− (0.3)0.045826 ≈ 0.218218 ≈ 0.22
P (Z > 0.22) = Area to the right of 0.22
Z
0.4129
0.22
0
P (p̂ > 0.31) = P (Z > 0.22) = 1− P (Z < 0.22) = 1− 0.5871
= 0.4129
P (Z < 0.22): in a z-table having area to the left of z, locate 0.2 in the left most...
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