Using normal probabilities in context | CAIE 9709 Probability & Statistics 1 Notes

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Probability & Statistics 1

Mathematics9709

Probability for one item becomes an expected count

one item

normal area = p

↓

many similar items

expected count = Np

fixed independent sample

Y ~ B(n, p)

Keep the roles separate: p describes one item, Np is an expected count, and B(n,p) models a new fixed set of independent selections.

\text{expected count}=N\times p

Here $p$ is the probability for one item and $N$ is the number of similar items. The product is a long-run expectation, not a guarantee for one particular group.

Percentage in the region: $100p\%$ .
Expected number in a group of size $N$ : $Np$ .
If the question asks how many people or items, finish with one sensible whole number.

Translate within d of the mean into two boundaries

“Within $d$ of the mean” means from $\mu-d$ to $\mu+d$ . The interval is centred on the mean.

Worked example

Expected days within one minute of the mean

Walking times have mean 14.8 minutes and standard deviation 1.5 minutes. Estimate how many of 225 days are within one minute of the mean.

z=\dfrac{1}{1.5}=0.6667

p=2\Phi(0.6667)-1

p=0.4950

225p=111.4

Answer

111\text{ days}

(9709/53/O/N/25 Q7(c))

Examiner note

The displayed probability is rounded, but multiply 225 by the stored calculator value. The published mark scheme accepts 111 or 112; give one integer and do not attach approximation wording to the answer.

Use the normal area as a binomial success probability

Suppose each selected item independently either meets the normal condition or does not. If $p$ is the normal probability that one item meets it, then the number $Y$ meeting it in a fixed sample of size $n$ has

Y\sim B(n,p)

Examiner note

Keep the stored value of

p

for the Binomial PD calculation. A direct binomial probability uses exact Binomial PD/CD. Only when the question explicitly says to use a normal or suitable approximation should you check both expected counts and apply a continuity correction.

Your turnIndependent check [6 marks]

Bottle fills are modelled by $X\sim N(500,4^2)$ . Estimate how many of 200 bottles exceed 504 ml. Then, for 12 independent bottles, find the probability that exactly 2 exceed 504 ml.

Show worked answer

z=\dfrac{504-500}{4}=1

p=0.158655\ldots

200p=31.731\ldots

So the expected count is 32. For the second part, let $Y$ count bottles above 504 ml.

Y\sim B(12,p)

With $q=1-p$ , the exact term is

P(Y=2)=\binom{12}{2}p^2q^{10}

Answer

32\text{ bottles}

P(Y=2)=0.295

Probability for one item becomes an expected count

one item

normal area = p

↓

many similar items

expected count = Np

fixed independent sample

Y ~ B(n, p)

Keep the roles separate: p describes one item, Np is an expected count, and B(n,p) models a new fixed set of independent selections.

\text{expected count}=N\times p

Here $p$ is the probability for one item and $N$ is the number of similar items. The product is a long-run expectation, not a guarantee for one particular group.

Percentage in the region: $100p\%$ .
Expected number in a group of size $N$ : $Np$ .
If the question asks how many people or items, finish with one sensible whole number.

Translate within d of the mean into two boundaries

“Within $d$ of the mean” means from $\mu-d$ to $\mu+d$ . The interval is centred on the mean.

Worked example

Expected days within one minute of the mean

Walking times have mean 14.8 minutes and standard deviation 1.5 minutes. Estimate how many of 225 days are within one minute of the mean.

z=\dfrac{1}{1.5}=0.6667

p=2\Phi(0.6667)-1

p=0.4950

225p=111.4

Answer

111\text{ days}

(9709/53/O/N/25 Q7(c))

Examiner note

Use the normal area as a binomial success probability

Y\sim B(n,p)

Examiner note

Keep the stored value of

p

Your turnIndependent check [6 marks]

Bottle fills are modelled by $X\sim N(500,4^2)$ . Estimate how many of 200 bottles exceed 504 ml. Then, for 12 independent bottles, find the probability that exactly 2 exceed 504 ml.

Show worked answer

z=\dfrac{504-500}{4}=1

p=0.158655\ldots

200p=31.731\ldots

So the expected count is 32. For the second part, let $Y$ count bottles above 504 ml.

Y\sim B(12,p)

With $q=1-p$ , the exact term is

P(Y=2)=\binom{12}{2}p^2q^{10}

Answer

32\text{ bottles}

P(Y=2)=0.295