Does this graph have 10 or 9 squares, 10 or 9 years, I think I might be going insane
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Problem 1, part A
X = number of times Roberto throws the baseball in the strike zone
X can take on the following values: 0, 1, 2, 3
p = 0.72 = probability of getting the ball in the strike zone (for any given independent trial)
n = 3 = sample size = number of times the baseball is thrown
Binomial Probabilities:
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 0) = (3 C 0)*(0.72)^(0)*(1-0.72)^(3-0)
P(X = 0) = (3 C 0)*(0.72)^(0)*(0.28)^(3)
P(X = 0) = (1)*(0.72)^(0)*(0.28)^3
P(X = 0) = (1)*(1)*(0.021952)
P(X = 0) = 0.021952
P(X = 0) = 0.022 <--- this value is added to the table (next to k = 0)
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 1) = (3 C 1)*(0.72)^(1)*(1-0.72)^(3-1)
P(X = 1) = (3 C 1)*(0.72)^(1)*(0.28)^(2)
P(X = 1) = (3)*(0.72)^(1)*(0.28)^2
P(X = 1) = (3)*(0.72)*(0.0784)
P(X = 1) = 0.169344
P(X = 1) = 0.169 <--- this value is added to the table (next to k = 1)
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 2) = (3 C 2)*(0.72)^(2)*(1-0.72)^(3-2)
P(X = 2) = (3 C 2)*(0.72)^(2)*(0.28)^(1)
P(X = 2) = (3)*(0.72)^(2)*(0.28)^1
P(X = 2) = (3)*(0.5184)*(0.28)
P(X = 2) = 0.435456
P(X = 2) = 0.435 <--- this value is added to the table (next to k = 2)
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 3) = (3 C 3)*(0.72)^(3)*(1-0.72)^(3-3)
P(X = 3) = (3 C 3)*(0.72)^(3)*(0.28)^(0)
P(X = 3) = (1)*(0.72)^(3)*(0.28)^0
P(X = 3) = (1)*(0.373248)*(1)
P(X = 3) = 0.373248
P(X = 3) = 0.373 <--- this value is added to the table (next to k = 3)
The table will look like what you see in the attached image
-----------------------------
Problem 1, part B
Refer to the table made in part A above. Add up the values in the second column, the P(X = k) column, that correspond to k values of 1 or larger. So basically everything but the first item which corresponds to k = 0
0.169+0.435+0.373 = 0.977
So the probability of at least one of the baseballs hits the strike zone is 0.977
=========================================
Problem 2
Convert each raw x score to a z score
Company A
z = (x-mu)/sigma
z = (260-276)/5.8
z = -2.759
----------------
Company B
z = (x-mu)/sigma
z = (260-252)/3.4
z = 2.353
----------------
The z scores are -2.759 and 2.353 for company A and company B respectively. The value -2.759 is further away from zero compared to 2.353, so company A has a lower chance of producing 260 nails. This is because company B has x = 260 closer to the mean (than compared to company A
X = number of times Roberto throws the baseball in the strike zone
X can take on the following values: 0, 1, 2, 3
p = 0.72 = probability of getting the ball in the strike zone (for any given independent trial)
n = 3 = sample size = number of times the baseball is thrown
Binomial Probabilities:
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 0) = (3 C 0)*(0.72)^(0)*(1-0.72)^(3-0)
P(X = 0) = (3 C 0)*(0.72)^(0)*(0.28)^(3)
P(X = 0) = (1)*(0.72)^(0)*(0.28)^3
P(X = 0) = (1)*(1)*(0.021952)
P(X = 0) = 0.021952
P(X = 0) = 0.022 <--- this value is added to the table (next to k = 0)
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 1) = (3 C 1)*(0.72)^(1)*(1-0.72)^(3-1)
P(X = 1) = (3 C 1)*(0.72)^(1)*(0.28)^(2)
P(X = 1) = (3)*(0.72)^(1)*(0.28)^2
P(X = 1) = (3)*(0.72)*(0.0784)
P(X = 1) = 0.169344
P(X = 1) = 0.169 <--- this value is added to the table (next to k = 1)
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 2) = (3 C 2)*(0.72)^(2)*(1-0.72)^(3-2)
P(X = 2) = (3 C 2)*(0.72)^(2)*(0.28)^(1)
P(X = 2) = (3)*(0.72)^(2)*(0.28)^1
P(X = 2) = (3)*(0.5184)*(0.28)
P(X = 2) = 0.435456
P(X = 2) = 0.435 <--- this value is added to the table (next to k = 2)
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 3) = (3 C 3)*(0.72)^(3)*(1-0.72)^(3-3)
P(X = 3) = (3 C 3)*(0.72)^(3)*(0.28)^(0)
P(X = 3) = (1)*(0.72)^(3)*(0.28)^0
P(X = 3) = (1)*(0.373248)*(1)
P(X = 3) = 0.373248
P(X = 3) = 0.373 <--- this value is added to the table (next to k = 3)
The table will look like what you see in the attached image
-----------------------------
Problem 1, part B
Refer to the table made in part A above. Add up the values in the second column, the P(X = k) column, that correspond to k values of 1 or larger. So basically everything but the first item which corresponds to k = 0
0.169+0.435+0.373 = 0.977
So the probability of at least one of the baseballs hits the strike zone is 0.977
=========================================
Problem 2
Convert each raw x score to a z score
Company A
z = (x-mu)/sigma
z = (260-276)/5.8
z = -2.759
----------------
Company B
z = (x-mu)/sigma
z = (260-252)/3.4
z = 2.353
----------------
The z scores are -2.759 and 2.353 for company A and company B respectively. The value -2.759 is further away from zero compared to 2.353, so company A has a lower chance of producing 260 nails. This is because company B has x = 260 closer to the mean (than compared to company A