Question
The weights, W grams, of bags of rice packaged in a factory can be modelled by a normal distribution with mean 204 grams and standard deviation 5 grams.
(a) A bag of rice is selected at random.
Find the probability that it weighs more than 210 grams.
According to this model, 80% of the bags of rice weigh between w grams and 210 grams.
(b) Find the probability that a randomly selected bag of rice weighs less than w grams.
(c) Find the value of w.
(d) Ten bags of rice are selected at random.
Find the probability that exactly one of the bags weighs less than w grams.
▶️Answer/Explanation
Answer:
(a) evidence of attempting to find correct area under normal curve
P(W>210) OR sketch
P(W>210) = 0.115069…
P(W>210) = 0.115
(b) recognizing P(W<w) = 1 – P(w<W<210)-P(W>210)
P(W<w) = 1 – 0.8 – 0.115069…
P(W<w) = 0.084930…
P(W<w) = 0.0849
(c) evidence of attempting to use inverse normal function
w =197.136…
w =197 (grams)
(d) recognition of binomial distribution
X ∼ B(10,0.0849302…)
P(X=1) = 0.382076…
P(X=1) = 0.382
Question
A bakery makes two types of muffins: chocolate muffins and banana muffins.
The weights, C grams, of the chocolate muffins are normally distributed with a mean of 62g and standard deviation of 2.9g.
(a) Find the probability that a randomly selected chocolate muffin weighs less than 61g.
(b) In a random selection of 12 chocolate muffins, find the probability that exactly 5 weigh less than 61g.
The weights, B grams, of the banana muffins are normally distributed with a mean of 68 g and standard deviation of 3.4g.
Each day 60% of the muffins made are chocolate.
On a particular day, a muffin is randomly selected from all those made at the bakery.
(c) (i) Find the probability that the randomly selected muffin weighs less than 61 g .
(ii) Given that a randomly selected muffin weighs less than 61g, find the probability that it is chocolate.
The machine that makes the chocolate muffins is adjusted so that the mean weight of the chocolate muffins remains the same but their standard deviation changes to σ g. The machine that makes the banana muffins is not adjusted. The probability that the weight of a randomly selected muffin from these machines is less than 61g is now 0.157.
(d) Find the value of σ .
Answer/Explanation
Ans:
(a) P(C< 61)
= 0.365112…
= 0.365
(b) recognition of binomial eg X ˜ B(12,0.365…)
P( x = 5) = 0.213666…
= 0.214
(c) (i) Let CM represent ‘chocolate muffin’ and BM represent ‘banana muffin’
P(B<61) = 0.0197555…
EITHER
P(CM) × P(C<61|CM) + P(BM) × P(B<61|BM) (or equivalent in words)
OR
tree diagram showing two ways to have a muffin weigh < 61
THEN
(0.6 × 0.365…) + (0.4 × 0.0197…)
= 0.226969…
= 0.227
(ii) recognizing conditional probability
Note: Recognition must be shown in context either in words or symbols, not just P (A|B) .
\(\frac{0.6\times 0.365112…}{0.22699…}\)
= 0.965183…
= 0.965
(d) METHOD 1
P(CM) × P(C<61|CM) + P(BM) × P(B<61|BM) = 0.157
(0.6 × P(C<61))+ (0.4 × 0.0197555….) = 0.157
P(C<61) = 0.248496….
attempt to solve for σ using GDC
Note: Award (M1) for a graph or table of values to show their P (C<61) with a variable standard deviation.
σ =1.47225…
σ = 1.47 (g)
METHOD 2
P(CM) × P(C<61|CM) + P(BM) × P(B<61|BM) = 0.157
(0.6 × P(C<61))+ (0.4 × 0.0197555….) = 0.157
P(C<61) = 0.248496….
use of inverse normal to find z score of their P(C< 61)
z = −0.679229…
correct substitution
\(\frac{61-62}{\sigma }\) = -0.679229…
σ =1.47225…
σ = 1.47 (g)
Question 8. [Maximum mark: 15]
The flight times, T minutes, between two cities can be modelled by a normal distribution with a mean of 75 minutes and a standard deviation of σ minutes.
(a) Given that 2% of the flight times are longer than 82 minutes, find the value of σ. [3]
(b) Find the probability that a randomly selected flight will have a flight time of more than 80 minutes. [2]
(c) Given that a flight between the two cities takes longer than 80 minutes, find the probability that it takes less than 82 minutes. [4]
On a particular day, there are 64 flights scheduled between these two cities.
(d) Find the expected number of flights that will have a flight time of more than 80 minutes. [3]
(e) Find the probability that more than 6 of the flights on this particular day will have a flight
time of more than 80 minutes.
Answer/Explanation
(a), use of inversse normal to find z-score
z=2.0537..
\(2.0537…=\frac{82-75}{\sigma }\)
\(\sigma=3.408401…\)
\(\sigma=3.41\)
(b) evidence of identifying the correct area under the normal curve
\(P\left ( T> 80 \right )=0.071193..\)
\(P\left ( T> 80 \right )=0.0712\)
(C) recognition that \( P\left ( 80 < T< 82\right )\) is required
\(P\left ( T< 82|T> 80 \right )=\frac{P\left ( 80< T< 82 \right )}{P\left ( T> 80 \right )}=\left ( \frac{0.51193….}{0.071193…} \right )\)
=7.19075..
=0.719
(d) recognition of binomial probability
\(X\sim B\left ( 64,0.071193.. \right )\) or \(E\left ( X \right )=64\times 0.071193\)
\(E\left ( X \right )=4.55353…\)
\(E\left ( X \right )=4.56\left ( flights \right )\)
(e)\( P\left ( X> 6 \right )=P\left ( X \geq 7\right )=1-P\left ( X\leq 6 \right )\)
=1-0.83088…
=0.0.1691196..
=0.169
Question
The probability distribution of a discrete random variable \(X\) is defined by
\(P(X = x) = cx(5 − x), x = 1, 2, 3, 4\).
- Find the value of \(c\).
- Find \(E(X)\).
Answer/Explanation
Ans:
- Using \(\sum P(X=x)=1\) ⇒ \(4c+6c+6c+6c+4c=1\) ⇒ \(20c=1\)
\(c=\frac{1}{20}(=0.05)\) - \(E(X)=(1×0.2)+(2×0.3)+(3×0.3)+(4×0.2)=2.5\)