Binomial Distribution
Normal Distribution
Discrete Probability
Random Maths
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100
EcoSmart is a company that produces light bulbs, and the probability of a bulb being defective is 0.05. What is the probability that out of 100 randomly selected bulbs, at most 3 will be defective?
Probability of Sucesses= 0.05
Number of Trials= 100
Number of Successes (x)= At most 3
BinomCDF (100, 0.05, 3)= 0.999
The probability that at most 3 bulbs will be effective is 0.999 or 99.9%
100
What is the standard deviation of this normal distribution X~N(10,4)
Standard Deviation= 2
100
Write three characteristics of a discrete probability distribution
1. 0≤P(x)≤1
2. ΣP(x)=1
3. Only p(x) for every value of x
100
Use the Fundamental Theorem of Calculus to evaluate
d/dx ∫_0^x(3+t)dx
3+x
100
Who said:
"No Luke, I am your Father"
200
In a survey, 80% of people prefer Starbucks over Dunkin. If 10 people were selected at random, what is the probability that exactly 7 of them would prefer Starbucks.
Probability of Success= 0.80
Number of trials= 10
X-Value= 7
BinomPDF(0.80, 10, 7)= 0.314
The probability that exactly 7 people would prefer Starbucks is 0.314 or 31.4%
200
What is the formula for calculating the z-score?
z= x - mean / standard deviation
200
A distribution is uniform from 2 to 8. What is the mean of this distribution?
5
200
∫_0^3(8-2x)dx
15
200
Who said:
"That's one small step for man, one giant leap for mankind"?
Neil Armstrong
300
Sai is a very good basketball player. He has a free throw success rate of 80%. If he attempts 10 free throws, what is the probability that he will make exactly 8 of them?
Success Rate= 0.80
Number of Trails= 10
Number of successes (x)= 8
BinomPDF(0.80,10,8)= 0.302
Therefore, the probability that Sai makes exactly 8 free throws is 0.302 or 30.2%
300
The height of all managers at Walmart is considered to be normally distributed, with a mean of 200cm and a standard deviation of 11cm. Determine the percentage of managers that are expected to be taller than 215cm?
Mean= 200
Standard Deviation= 11
Upper Bound= Infinity
Lower Bound= 215
NormCDF(200,11,215,100000)= 0.0863
1-0.0863=0.137
The percentage of the managers that are expected to be taller than 215cm is 13.7%
300
Find the mean number of breakdowns per week for this machine
x P(x)
0 .15
1 .20
2 .35
3 .30
1.8
300
Solve:
log2 x +log2(x-7)=3
x = 8
300
Who delivered the speech beginning,
"We choose to go to the Moon"?
John F Kennedy
400
In a hypothetical scenario, if the probability of success in each trial is 0.3 and the number of trials is 10, what is the probability of getting exactly 3 successes?
The probability of success= is 0.3
Number of trials=10
Number of Successes (x)= 3
BinomPDF (0.3,10,3)= 0.267
The probability of getting exactly 3 successes is 0.267 or about 26.7%
400
Most colleges require SAT scores in order to get accepted. For Drexel University, scores on the SAT are roughly normally distributed with a mean of 1200 and a standard deviation of 100. What is the probability of an individual scoring above 1000 on the SAT?
Mean= 1200
Standard Deviation= 100
We first need to calculate the z-score
Z=(1000-1200)/100= -2
Probability(X > 1000) = 1 - normCDF(1000, infinity, 1200, 100)
1 - 0.9772 ≈ 0.0228
The probability of an individual scoring above 1000 on the SAT is approximately 0.0228 or 2.28%
400
Let x be the number of heads obtained in two tosses of a fair coin. Calculate the mean and standard deviation of x
μ=1.00
σ=.707
400
Solve:
[log5(x)]2-+7log5(x)-30=0
x=5-10 or 125
400
Who gave the speech "Their Finest Hour" during World War II?
Winston Churchill
500
A fair coin is tossed 8 times. What is the probability of getting at least 6 heads?
To find the probability of getting at least 6 heads, we need to calculate the probabilities of getting exactly 6, 7, and 8 heads, and then add them together.
P(X ≥ 6) = P(X = 6) + P(X = 7) + P(X = 8)
Success Rate= 0.50
Number of Trials= 8
BinomPDF(0.5,8,6)= 0.109
BinomPDF(0.5,8,7)= 0.0625
BinomPDF(0.5,8,8)=0.0039
Add all the probability for final result
0.109+0.0625+0.0039= 0.1754
The probability of getting at least 6 heads is 0.1754 or 17.54%
500
AMC Movies has done a study to see how much money their customers are willing to spend on snacks like popcorn and soda. It is found that the spending pattern is normally distributed with a mean of $20 and a standard deviation of $5. What percentage of customers will spend less than $10 on snacks?
Mean= 20
Standard Deviation= 5
X=10
Z Score= 10-20/ 5= -2
Corresponding Z-Score value based on standard normal distribution table is 0.02275
The percentage of customers that will spend less than $10 on snacks is about 2.28%
500
According to the most recent data from the Insurance Research Council, 16.1% of motorists in the United States were uninsured in 2010. Suppose that current 16.1% of motorists in the United States are uninsured. Suppose that two motorists are selected at random. Let x denote the number of motorists in this sample of two who are uninsured. Construct the probability distribution table of x.
X P(x)
0 .7039
1 .2702
2 .0259
500
Find Var(X)
20/9 = 2.22222
500
“The sum of the square roots of any two sides of an isosceles triangle is equal to the square root of the remaining side.”