Edexcel S1 2023 June — Question 7 14 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
Year2023
SessionJune
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNormal Distribution
TypeProbability calculation plus find unknown boundary
DifficultyStandard +0.3 This is a standard S1 normal distribution question requiring routine standardisation, use of tables, and solving simultaneous equations with z-scores. Part (a) is straightforward table lookup, part (b) involves basic probability with independence, and part (c) requires setting up two equations from given probabilities—all standard techniques with no novel insight needed, making it slightly easier than average.
Spec2.04c Calculate binomial probabilities2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation
A machine squeezes apples to extract their juice. The volume of juice, \(J\) ml, extracted from 1 kg of apples is modelled by a normal distribution with mean \(\mu\) and standard deviation \(\sigma\) Given that \(\mu = 500\) and \(\sigma = 25\) use standardisation to
    1. show that P\((J > 510) = 0.3446\) [2]
    2. calculate the value of \(d\) such that P\((J > d) = 0.9192\) [3]
Zen randomly selects 5 bags each containing 1 kg of apples and records the volume of juice extracted from each bag of apples.
  1. Calculate the probability that each of the 5 bags of apples produce less than 510ml of juice. [2]
Following adjustments to the machine, the volume of juice, \(R\) ml, extracted from 1 kg of apples is such that \(\mu = 520\) and \(\sigma = k\) Given that P\((R < r) = 0.15\) and P\((R > 3r - 800) = 0.005\)
  1. find the value of \(r\) and the value of \(k\) [7]
Question 7:
AnswerMarks
7(a)(i) 510−500
P(J 510) = P Z   or P(Z > 0.4)
AnswerMarks
 25 M1
= 1 – 0.6554 0.3446 *A1*
(2)
AnswerMarks
(ii)d −500
=−1.4(calc –1.3997…)
AnswerMarks
25M1B1
d = 465 (calc 465.007)dA1
(3)
AnswerMarks Guidance
(b)( 1−0.3446 )5 M1
= 0.1209… awrt 0.121A1
(2)
AnswerMarks
(c)r−520
=−1.0364
AnswerMarks
kM1A1
3r−800−520
=2.5758
AnswerMarks
kM1A1
r−520 3r−1320
−240=(3−"1.0364k")−"2.5758"k or = oe
AnswerMarks
"−1.0364" "2.5758"ddM1
k =42.216... awrt 42A1
r =476.246... awrt 476dA1
(7)
AnswerMarks
NotesTotal 14
(a)(i)M1 for standardising using 500 and 25. Allow for 0.4
A1* M1 must be awarded. For 1 – 0.6554 = 0.3446 or using calc 0.34457… = 0.3446 or better
AnswerMarks
(ii)M1 correct standardisation using 500 and 25 equated to a z value where1 z 2
500−d
B1 correct expression with compatible signs eg =1.4 (calc 1.3997…) or allow incompatible
25
signs with 500−("535"−500)
510−d
SC =1.4 (calc 1.3997…) can get M0B1A0
25
dA1 dependent on M1 awarded for 465 or 465.007...
AnswerMarks
(b)M1 for (p)5 where 0 < p < 1
A1 awrt 0.121
AnswerMarks Guidance
(c)r−520
M1 = z value wherez > 1
k
r−520
1st A1 =awrt −1.0364(calc 1.036433...) (signs must be compatible)
k
3r−800−520
AnswerMarks Guidance
2nd M1 = z value wherez > 2
k
3r−800−520
2nd A1 =awrt 2.5758 (calc 2.5758293…) (signs must be compatible)
k
3rd M1 (dep on both Ms) for forming a correct equation in k or r only using their z values. ISW once
3(−1.0364k+520)−800−520
correct equation seen eg –5.685k = –240 or =2.5758 Implied by r
k
and k correct
3rdA1 for awrt 42
4th A1 for awrt 476 Must come from equations with compatible signs
NB awrt 476 and awrt 42 does not mean full marks. They could get M1A0M1A0 M1A1A1 if they do
not have accurate z values
PMT
Pearson Education Limited. Registered company number 872828
with its registered office at 80 Strand, London, WC2R 0RL, United Kingdom
Question 7:
--- 7(a)(i) ---
7(a)(i) |  510−500
P(J 510) = P Z   or P(Z > 0.4)
 25  | M1
= 1 – 0.6554 0.3446 * | A1*
(2)
(ii) | d −500
=−1.4(calc –1.3997…)
25 | M1B1
d = 465 (calc 465.007) | dA1
(3)
(b) | ( 1−0.3446 )5 | M1
= 0.1209… awrt 0.121 | A1
(2)
(c) | r−520
=−1.0364
k | M1A1
3r−800−520
=2.5758
k | M1A1
r−520 3r−1320
−240=(3−"1.0364k")−"2.5758"k or = oe
"−1.0364" "2.5758" | ddM1
k =42.216... awrt 42 | A1
r =476.246... awrt 476 | dA1
(7)
Notes | Total 14
(a)(i) | M1 for standardising using 500 and 25. Allow for 0.4
A1* M1 must be awarded. For 1 – 0.6554 = 0.3446 or using calc 0.34457… = 0.3446 or better
(ii) | M1 correct standardisation using 500 and 25 equated to a z value where1 z 2
500−d
B1 correct expression with compatible signs eg =1.4 (calc 1.3997…) or allow incompatible
25
signs with 500−("535"−500)
510−d
SC =1.4 (calc 1.3997…) can get M0B1A0
25
dA1 dependent on M1 awarded for 465 or 465.007...
(b) | M1 for (p)5 where 0 < p < 1
A1 awrt 0.121
(c) | r−520
M1 = z value where |z| > 1
k
r−520
1st A1 =awrt −1.0364(calc 1.036433...) (signs must be compatible)
k
3r−800−520
2nd M1 = z value where |z| > 2
k
3r−800−520
2nd A1 =awrt 2.5758 (calc 2.5758293…) (signs must be compatible)
k
3rd M1 (dep on both Ms) for forming a correct equation in k or r only using their z values. ISW once
3(−1.0364k+520)−800−520
correct equation seen eg –5.685k = –240 or =2.5758 Implied by r
k
and k correct
3rdA1 for awrt 42
4th A1 for awrt 476 Must come from equations with compatible signs
NB awrt 476 and awrt 42 does not mean full marks. They could get M1A0M1A0 M1A1A1 if they do
not have accurate z values
PMT
Pearson Education Limited. Registered company number 872828
with its registered office at 80 Strand, London, WC2R 0RL, United Kingdom
A machine squeezes apples to extract their juice. The volume of juice, $J$ ml, extracted from 1 kg of apples is modelled by a normal distribution with mean $\mu$ and standard deviation $\sigma$

Given that $\mu = 500$ and $\sigma = 25$ use standardisation to

\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item show that P$(J > 510) = 0.3446$ [2]
\item calculate the value of $d$ such that P$(J > d) = 0.9192$ [3]
\end{enumerate}
\end{enumerate}

Zen randomly selects 5 bags each containing 1 kg of apples and records the volume of juice extracted from each bag of apples.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Calculate the probability that each of the 5 bags of apples produce less than 510ml of juice. [2]
\end{enumerate}

Following adjustments to the machine, the volume of juice, $R$ ml, extracted from 1 kg of apples is such that $\mu = 520$ and $\sigma = k$

Given that P$(R < r) = 0.15$ and P$(R > 3r - 800) = 0.005$

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item find the value of $r$ and the value of $k$ [7]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S1 2023 Q7 [14]}}
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