A container is chosen at random. Find the probability that it is underfilled.

The company decides that the probability of a container being underfilled should be reduced to $0.02$. It decreases the standard deviation to $\sigma$ and leaves the mean unchanged.

Find $\sigma$.

The company changes to the new standard deviation, $\sigma$, and leaves the mean unchanged.

A container is chosen at random for inspection. It passes inspection if its volume of yogurt is between $250$ and $271$ ml.

(i)     Find the probability that it passes inspection.

(ii)     Given that the container is not underfilled, find the probability that it passes inspection.

A sample of $50$ containers is chosen at random. Find the probability that $48$ or more of the containers pass inspection.

$0.0477903$

probability $= 0.0478$     A2     N2

[2 marks]

${\text{P}}({\text{volume}} < 250) = 0.02$     (M1)

$z =  - 2.05374\;\;\;$(may be seen in equation)     A1

attempt to set up equation with $z$     (M1)

eg$\;\;\;\frac{{\mu  - 260}}{\sigma } = z,{\text{ }}260 - 2.05(\sigma ) = 250$

$4.86914$

$\sigma  = 4.87{\text{ (ml)}}$     A1     N3

[4 marks]

(i)     $0.968062$

${\text{P}}(250 < {\text{Vol}} < 271) = 0.968$     A2     N2

(ii)     recognizing conditional probability (seen anywhere, including in correct working)     R1

eg$\;\;\;{\text{P}}(A|B),{\text{ }}\frac{{{\text{P}}(A \cap B)}}{{{\text{P}}(B)}},{\text{ P}}(A \cap B) = {\text{P}}(A|B){\text{P}}(B)$

correct value or expression for $P$ (not underfilled)     (A1)

eg$\;\;\;0.98,1 - 0.02,{\text{ }}1 - {\text{P}}(X < 250)$

probability $= \frac{{0.968}}{{0.98}}$     A1

$0.987818$

probability $= 0.988$     A1     N2

[6 marks]

METHOD 1

evidence of recognizing binomial distribution (seen anywhere)     (M1)

eg$\;\;\;X\;\;\;{\text{B}}(50,{\text{ }}0.968),{\text{ binomial cdf, }}p = 0.968,{\text{ }}r = 47$

${\text{P}}(X \le 47$) = 0.214106\)     (A1)

evidence of using complement     (M1)

eg$\;\;\;1 - {\text{P}}(X \le 47$)

$0.785894$

probability $= 0.786$     A1     N3

METHOD 2

evidence of recognizing binomial distribution (seen anywhere)     (M1)

eg$\;\;\;X\;\;\;{\text{B}}(50,{\text{ }}0.968),{\text{ binomial cdf, }}p = 0.968,{\text{ }}r = 47$

${\text{P(not pass)}} = 1 - {\text{P(pass)}} = 0.0319378$     (A1)

evidence of attempt to find $P$ ($2$ or fewer fail)     (M1)

eg$\;\;\;$$0$, $1$, or $2$ not pass, ${\text{B}}(50,{\text{ }}2)$

$0.785894$

probability $= 0.786$     A1     N3

METHOD 3

evidence of recognizing binomial distribution (seen anywhere)     (M1)

eg$\;\;\;X\;\;\;{\text{B}}(50,{\text{ }}0.968),{\text{ binomial cdf, }}p = 0.968,{\text{ }}r = 47$

evidence of summing probabilities     (M1)

eg$\;\;\;{\text{P}}(X = 48) + {\text{P}}(X = 49) + {\text{P}}(X = 50)$

correct working

eg$\;\;\;0.263088 + 0.325488 + 0.197317$     (A1)

$0.785894$

probability $= 0.786$     A1     N3

[4 marks]

Total [16 marks]

This question saw many candidates competently using their GDCs to obtain required values, although a surprising number in part (b) chose to use an inefficient ‘guess and check’ method to try and obtain the standard deviation. Those using a correct approach often used a rounded z-score to find $\sigma$ leading to an inaccurate final answer. In part (c), some candidates did not recognize or understand how to apply the given condition. In part (d), the binomial distribution, although often recognized, was not applied successfully.

This question saw many candidates competently using their GDCs to obtain required values, although a surprising number in part (b) chose to use an inefficient ‘guess and check’ method to try and obtain the standard deviation. Those using a correct approach often used a rounded z-score to find $\sigma$ leading to an inaccurate final answer. In part (c), some candidates did not recognize or understand how to apply the given condition. In part (d), the binomial distribution, although often recognized, was not applied successfully.

This question saw many candidates competently using their GDCs to obtain required values, although a surprising number in part (b) chose to use an inefficient ‘guess and check’ method to try and obtain the standard deviation. Those using a correct approach often used a rounded z-score to find $\sigma$ leading to an inaccurate final answer. In part (c), some candidates did not recognize or understand how to apply the given condition. In part (d), the binomial distribution, although often recognized, was not applied successfully.

This question saw many candidates competently using their GDCs to obtain required values, although a surprising number in part (b) chose to use an inefficient ‘guess and check’ method to try and obtain the standard deviation. Those using a correct approach often used a rounded z-score to find $\sigma$ leading to an inaccurate final answer. In part (c), some candidates did not recognize or understand how to apply the given condition. In part (d), the binomial distribution, although often recognized, was not applied successfully.