SME
Grade 12 Math · Problem type
Mean and spread of a binomial count
Use this for “In 40 trials with p = 0.25, find the mean and standard deviation of the count” (up to 40 trials).
Chart
Binomial bars: n = 40 trials, each a success with chance p = 0.25.
μ = n × p = 40 × 0.25 = 10
σ = √(n × p × (1 − p)) = 2.7386
This is an example. Type your own values to start a new problem.
μ
=
n
×
p
σ
=
×
×
(1
−
)
Formulas
μ
=
n
×
p
10 = 40 × 0.25
σ
=
√(
n
×
p
×
(1
−
p
))
2.7386 = √(40 × 0.25 × (1 − 0.25))
Assumptions
•
n independent trials, each a success with the same chance p; X counts the successes.
•
X has mean np and standard deviation √(np(1 − p)).
•
X is close to normal when np ≥ 10 and n(1 − p) ≥ 10.
Step-by-step
WE KNOW
n = 40, p = 0.25
FIND
mean count (μ), standard deviation of the count (σ)
Step 1 · Find mean count (μ)
Use μ = n × p
On
average,
the
share
p
of
the
n
trials
succeed.
μ
=
40
×
0.25
μ
=
10
Step 2 · Find standard deviation of the count (σ)
Use σ = √(n × p × (1 − p))
Each
trial
adds
p
(1
−
p
)
to
the
variance;
the
root
is
the
spread
of
the
count.
σ
=
√(40
×
0.25
×
(1
−
0.25))
σ
=
√
40 × 0.25 × 0.75
σ
=
√
10 × 0.75
σ
=
√
7.5
σ
=
2.7386
Check
Put the numbers back in. Both sides should match.
10 = 40 × 0.25 ✓
2.7386 ≈ √(40 × 0.25 × (1 − 0.25)) ✓
← Previous lesson
Sampling distribution of a proportion
Next lesson →
The central limit theorem, simulated
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Refresh
Related lessons
Sampling distributions and the central limit theorem
Main lesson
Sampling distribution of a proportion
Use this for “40% of voters agree. In a sample of 150, what is P(p̂ > 0.46)?”
The central limit theorem, simulated
Use this for “Wait times are skewed right with mean 4 minutes. How are the means of samples of 30 spread?”
Sample means and sums with any σ
Use this for “Bags of rice weigh 2 kg on average with σ = 0.12 kg. Find the chance that 36 bags average over 2.03 kg, or weigh over 73.08 kg in all.”