Tuesday, 10/8
➢ Binomial probability distribution example-a waitress notices the probability of a
customer leaving a tip is 0.5. With six diners, make a binomial probability distribution
showing the probabilities that 0, 1, 2, 3, 4, 5, and all 6 diners will leave a tip.
→ N = 6, p = 0.5, q = 0.5
→ Where n is the number of trials, p is the probability of success, and q is the
probability of failure(1-p)
r P(r)
0 0.016
1 0.094
2 0.234
3 0.312
4 0.234
5 0.094
6 0.016
0.3
0.2
0.1
0 1 2 3 4 5
→ 6
, → Formula:
→ If p = 0.20 for example, graph will be skewed right; if p>0.5, the graph will be
skewed left
➢ µ = np: expected number of = successes for r
➢ δ = √npq
➢ Find mean(µ) and standard deviation(δ) for data from earlier example
→ µ = 6 * 0.5 = 3
→ δ = √6 * 0.5 * 0.5 = √1.5 = 1.22
Guided Exercise
➢ Tashika makes 80% of free throw attempts(probability = 0.80)
→ Out of 6 free throws, what is the expected number(µ)of free throws Tashika will
make?
→ For six trials, what is the standard deviation?
→ Mean: µ = 6 * 0.8 = 4.8
→ Standard deviation: δ = √6 * 0.8 * 0.2 = √0.96 = 0.98
→ So, we expect Tashika to make 5 out of 6 free throws(rounding up from 4.8)
Guided Exercise 2
➢ Binomial probability distribution example-a waitress notices the probability of a
customer leaving a tip is 0.5. With six diners, make a binomial probability distribution
showing the probabilities that 0, 1, 2, 3, 4, 5, and all 6 diners will leave a tip.
→ N = 6, p = 0.5, q = 0.5
→ Where n is the number of trials, p is the probability of success, and q is the
probability of failure(1-p)
r P(r)
0 0.016
1 0.094
2 0.234
3 0.312
4 0.234
5 0.094
6 0.016
0.3
0.2
0.1
0 1 2 3 4 5
→ 6
, → Formula:
→ If p = 0.20 for example, graph will be skewed right; if p>0.5, the graph will be
skewed left
➢ µ = np: expected number of = successes for r
➢ δ = √npq
➢ Find mean(µ) and standard deviation(δ) for data from earlier example
→ µ = 6 * 0.5 = 3
→ δ = √6 * 0.5 * 0.5 = √1.5 = 1.22
Guided Exercise
➢ Tashika makes 80% of free throw attempts(probability = 0.80)
→ Out of 6 free throws, what is the expected number(µ)of free throws Tashika will
make?
→ For six trials, what is the standard deviation?
→ Mean: µ = 6 * 0.8 = 4.8
→ Standard deviation: δ = √6 * 0.8 * 0.2 = √0.96 = 0.98
→ So, we expect Tashika to make 5 out of 6 free throws(rounding up from 4.8)
Guided Exercise 2
