Which table is a probability distribution table ?

Mathematics

2 answers:

Aleonysh [2.5K]4 years ago

7 0

Answer:

the 2 one

Step-by-step explanation:

Sveta_85 [38]4 years ago

4 0

Answer:

Third table.

Step-by-step explanation:

A probabilities is the quotient of the number of favourable outcomes and the total number of possible outcomes. From its definition, we can deduct that the numbers are always positive, and the denominator is always major than the numerator. So, every probability is restricted from 0 to 1, that the range of probabilities results, otherwise, if the number is negative or more than 1, that's not a probabilities.

So, the first and fourth are not a probability distribution because they have negative "probabilities". The second table is not a probability distribution either, because the sum of all is more than 1, which is not possible.

Therefore, the right answer is third table because:

0.6 + 0.05 + 0.05 + 0.3 = 1; and all numbers are positive.

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A. 16%

k0ka [10]

I'd say it would be 24% but i may be wrong

8 0

4 years ago

Read 2 more answers

Blurry but an you help me

Korolek [52]

Answer:

D)5/11

Step-by-step explanation:

So yeah there's 22 balls in total!

Then for the red balls, there are the balls numbered evenly: 6, 8, 10, 12, 14. So there's 5 even balls for red.

Then for the green balls: 8, 10, 12, 14, 16, there are also 5!

5+5=10

10/22

=5/11

The answer is D!

Hoped I helped (P.S. I love math)

8 0

3 years ago

Read 2 more answers

This is what I need help on....

OleMash [197]

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