Tommy Wait, a minor league baseball pitcher, is notorious for taking an excessive amount of time between pitches. In fact, his time between pitches are normally distributed with a mean of seconds and a standard deviation of seconds. What percentage of his times between pitches are longer than seconds

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

Tommy Wait, a minor league baseball pitcher, is notorious for taking an excessive amount of time between pitches. In fact, his time between pitches are normally distributed with a mean of 29 seconds and a standard deviation of 2.1 seconds. What percentage of his times between pitches are longer than 31 seconds ?

Given Information:  

Mean pitching time = μ = 29 seconds

Standard deviation of pitching time = σ = 2.1 seconds

Required Information:  

P(X > 31) = ?

Step-by-step explanation:

What is Normal Distribution?

We are given a Normal Distribution, which is a continuous probability distribution and is symmetrical around the mean. The shape of this distribution is like a bell curve and most of the data is clustered around the mean. The area under this bell shaped curve represents the probability.  

We want to find out the probability that what percentage of his times between pitches are longer than 31 seconds.

The z-score corresponding to 0.95 is 0.8289

Therefore, 17.11% of his times between pitches are longer than 31 seconds.

How to use z-table?

Step 1:

In the z-table, find the two-digit number on the left side corresponding to your z-score. (e.g 0.9 1.4, 2.2, 0.5 etc.)

Step 2:

Then look up at the top of z-table to find the remaining decimal point in the range of 0.00 to 0.09. (e.g. if you are looking for 0.95 then go for 0.05 column)

Step 3:

Finally, find the corresponding probability from the z-table at the intersection of step 1 and step 2.

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!

step-by-step explanation:

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