Mathematics · May / June 2021 · Paper 61 · Question 4

YMA01

As printed in the paper

The question as text

4. Kris works in the mailroom of a large company and is responsible for all the letters sent by the company. The weights of letters sent by the company, W grams, have a normal distribution with mean 165 g and standard deviation 35 g.

Read automatically from the paper; the printed page above is exact.

Show the answer and mark scheme

4(a)[3]
P(W < 120) =
P
Z
<
35
= P(Z < – 1.2857…) = 1 – 0.9015 or 1 – 0.9007285…
= 0.09927…= awrt 0.0985~0.0994
x −
165
(b) e.g. P(W > x) = 1
3 gives
0.43
(calculator 0.430727…)
= ±
35
Limits 149.9245… to 180.0754… awrt 150 to 180
P("180"
200)
<
W
<
(c) P(W < 200 | W > “180”) or
P(
W
>
"180") or
1
3
0.8413(44739...) −
2
=
3
1
3
= 0.52403… (0.523~0.5264)
1
1
1; 3!
4(b)[4]
P(W < 120) =
P
Z
<
35
= P(Z < – 1.2857…) = 1 – 0.9015 or 1 – 0.9007285…
= 0.09927…= awrt 0.0985~0.0994
x −
165
(b) e.g. P(W > x) = 1
3 gives
0.43
(calculator 0.430727…)
= ±
35
Limits 149.9245… to 180.0754… awrt 150 to 180
P("180"
200)
<
W
<
(c) P(W < 200 | W > “180”) or
P(
W
>
"180") or
1
3
0.8413(44739...) −
2
=
3
1
3
= 0.52403… (0.523~0.5264)
1
1
1; 3!
×
×
×
3
3
3
= 2
9
Notes
(a) 1st M1 for standardising with 120 (allow 210), 165 and 35. Accept +
2nd M1 for attempting 1 – p [where 0.85 < p < 0.95]
A1 for awrt 0.0985~0.0994 (Correct ans only 3/3)
(b) M1 for standardising with x (o.e.) 165 and 35 and setting equal to a z value, 0.4 < |z| < 0.5
(Accept 165
−
x
= ± where 0.4 < |z| < 0.5)
z
35
B1 for use of z = 0.43 or better We must see 0.43 or better.
1st A1 for lower limit of awrt 150
2nd A1 for upper limit of awrt 180
SC A0A1 for two limits symmetrically placed about 165 provided M1 scored
NB: correct answers with no working can score M1B0A1A1
(c) M1 for a correct probability statement (either form) ft their 180 or a correct ratio
1st A1 for a correct numerator (awrt 0.175)
2nd A1 for an answer in the range awrt 0.523~0.5264 (use of 180 gives 0.5263869…)
(d) 1st M1 for ( )
3
1
(or equivalent)
3
2nd M1 for p 3!
×
(or equivalent) where 0 < p < 1
6
for 2
9 or any exact equivalent
A1
5. (a) {E(X) = } 2
−
a
−
b
+ × +
0
c
b
+
4
a
or 2a
{ 2a = 0.5 so } a = 0.25
(b) {
(
)
E(
X
2
) }
=
−
2
2
×
a
+ −
( 1)
2
×
b
+
0
+
1
2
×
b
+
4
2
×
a
or 20a + 2b (o.e.)
{Var(X) =}"20a + 2b" – 0.52
20a + 2b – 0.25 = 5.01 (o.e.) e.g. "4.75" + 2b = 5.01
{ 2b = 0.26 so } b = 0.13
{Use of sum of probs = 1 to calculate a 2nd value} c = 0.24
(c)(i)
{E(Y) = 5 – 8×0.5 } = 1
(ii)
{Var(Y) =}
( 8)
−
2
×
5.01
= 320.64 awrt 321
(d)
4
X
2
>
5 8
−
X
(
2
1 2
)(
5
)
0
0.5
X
−
X
+
>
⇒
X
>
So need X = 1 or 4 or probability of a + b
= 0.38
Notes
(a) M1
for any correct expression for E(X) in terms of a (or a, b, c)
A1
for a = 0.25
(b) 1st M1
for attempt at an expression for E(X 2) with at least 3 correct nonzero terms
2nd M1
for a correct expression for Var(X) eg"18a – c + 1" – 0.52 Allow with their value
of a substituted
1st A1 for a correct equation for b (or possibly c) eg"18a – c + 1" – 0.52 = 5.01 Allow
with their value of a substituted
2nd A1 for either b = 0.13 or c = 0.24
3rd A1ft for using c = 1 – 2×"0.25" – 2×"0.13" or b = (1 – 2×"0.25" –"0.24" ) ÷2 to gain
the correct ft answer for their 2nd value
(c) B1
for {E(Y) =} 1
M1
for correct use of Var(aX + b) = a2 Var (X )
A1
for awrt 321
(d) 1st M1 for correct quadratic inequality (may be inside prob statement) or table of values
2nd M1 for an attempt to solve or identifying correct X values
1st A1 for X > 0.5 [ may also have X < – 2.5 ]
3rd M1 for realising need X = 1 and 4 only or answer of their (a + b)
2nd A1 for 0.38 (or exact equivalent) only (correct ans only 5/5)
2
{
}
23.7
4(c)[3]
P(W < 120) =
P
Z
<
35
= P(Z < – 1.2857…) = 1 – 0.9015 or 1 – 0.9007285…
= 0.09927…= awrt 0.0985~0.0994
x −
165
(b) e.g. P(W > x) = 1
3 gives
0.43
(calculator 0.430727…)
= ±
35
Limits 149.9245… to 180.0754… awrt 150 to 180
P("180"
200)
<
W
<
(c) P(W < 200 | W > “180”) or
P(
W
>
"180") or
1
3
0.8413(44739...) −
2
=
3
1
3
= 0.52403… (0.523~0.5264)
1
1
1; 3!
×
×
×
3
3
3
= 2
9
Notes
(a) 1st M1 for standardising with 120 (allow 210), 165 and 35. Accept +
2nd M1 for attempting 1 – p [where 0.85 < p < 0.95]
A1 for awrt 0.0985~0.0994 (Correct ans only 3/3)
(b) M1 for standardising with x (o.e.) 165 and 35 and setting equal to a z value, 0.4 < |z| < 0.5
(Accept 165
−
x
= ± where 0.4 < |z| < 0.5)
z
35
B1 for use of z = 0.43 or better We must see 0.43 or better.
1st A1 for lower limit of awrt 150
2nd A1 for upper limit of awrt 180
SC A0A1 for two limits symmetrically placed about 165 provided M1 scored
NB: correct answers with no working can score M1B0A1A1
(c) M1 for a correct probability statement (either form) ft their 180 or a correct ratio
1st A1 for a correct numerator (awrt 0.175)
2nd A1 for an answer in the range awrt 0.523~0.5264 (use of 180 gives 0.5263869…)
(d) 1st M1 for ( )
3
1
(or equivalent)
3
2nd M1 for p 3!
×
(or equivalent) where 0 < p < 1
6
for 2
9 or any exact equivalent
A1
5. (a) {E(X) = } 2
−
a
−
b
+ × +
0
c
b
+
4
a
or 2a
{ 2a = 0.5 so } a = 0.25
(b) {
(
)
E(
X
2
) }
=
−
2
2
×
a
+ −
( 1)
2
×
b
+
0
+
1
2
×
b
+
4
2
×
a
or 20a + 2b (o.e.)
{Var(X) =}"20a + 2b" – 0.52
20a + 2b – 0.25 = 5.01 (o.e.) e.g. "4.75" + 2b = 5.01
{ 2b = 0.26 so } b = 0.13
{Use of sum of probs = 1 to calculate a 2nd value} c = 0.24
(c)(i)
{E(Y) = 5 – 8×0.5 } = 1
(ii)
{Var(Y) =}
( 8)
−
2
×
5.01
= 320.64 awrt 321
(d)
4
X
2
>
5 8
−
X
(
2
1 2
)(
5
)
0
0.5
X
−
X
+
>
⇒
X
>
So need X = 1 or 4 or probability of a + b
= 0.38
Notes
(a) M1
for any correct expression for E(X) in terms of a (or a, b, c)
A1
for a = 0.25
(b) 1st M1
for attempt at an expression for E(X 2) with at least 3 correct nonzero terms
2nd M1
for a correct expression for Var(X) eg"18a – c + 1" – 0.52 Allow with their value
of a substituted
1st A1 for a correct equation for b (or possibly c) eg"18a – c + 1" – 0.52 = 5.01 Allow
with their value of a substituted
2nd A1 for either b = 0.13 or c = 0.24
3rd A1ft for using c = 1 – 2×"0.25" – 2×"0.13" or b = (1 – 2×"0.25" –"0.24" ) ÷2 to gain
the correct ft answer for their 2nd value
(c) B1
for {E(Y) =} 1
M1
for correct use of Var(aX + b) = a2 Var (X )
A1
for awrt 321
(d) 1st M1 for correct quadratic inequality (may be inside prob statement) or table of values
2nd M1 for an attempt to solve or identifying correct X values
1st A1 for X > 0.5 [ may also have X < – 2.5 ]
3rd M1 for realising need X = 1 and 4 only or answer of their (a + b)
2nd A1 for 0.38 (or exact equivalent) only (correct ans only 5/5)
2
{
}
23.7
4(d)[3]
×
×
×
3
3
3
= 2
9
Notes
(a) 1st M1 for standardising with 120 (allow 210), 165 and 35. Accept +
2nd M1 for attempting 1 – p [where 0.85 < p < 0.95]
A1 for awrt 0.0985~0.0994 (Correct ans only 3/3)
(b) M1 for standardising with x (o.e.) 165 and 35 and setting equal to a z value, 0.4 < |z| < 0.5
(Accept 165
−
x
= ± where 0.4 < |z| < 0.5)
z
35
B1 for use of z = 0.43 or better We must see 0.43 or better.
1st A1 for lower limit of awrt 150
2nd A1 for upper limit of awrt 180
SC A0A1 for two limits symmetrically placed about 165 provided M1 scored
NB: correct answers with no working can score M1B0A1A1
(c) M1 for a correct probability statement (either form) ft their 180 or a correct ratio
1st A1 for a correct numerator (awrt 0.175)
2nd A1 for an answer in the range awrt 0.523~0.5264 (use of 180 gives 0.5263869…)
(d) 1st M1 for ( )
3
1
(or equivalent)
3
2nd M1 for p 3!
×
(or equivalent) where 0 < p < 1
6
for 2
9 or any exact equivalent
A1
5. (a) {E(X) = } 2
−
a
−
b
+ × +
0
c
b
+
4
a
or 2a
{ 2a = 0.5 so } a = 0.25
(b) {
(
)
E(
X
2
) }
=
−
2
2
×
a
+ −
( 1)
2
×
b
+
0
+
1
2
×
b
+
4
2
×
a
or 20a + 2b (o.e.)
{Var(X) =}"20a + 2b" – 0.52
20a + 2b – 0.25 = 5.01 (o.e.) e.g. "4.75" + 2b = 5.01
{ 2b = 0.26 so } b = 0.13
{Use of sum of probs = 1 to calculate a 2nd value} c = 0.24
(c)(i)
{E(Y) = 5 – 8×0.5 } = 1
(ii)
{Var(Y) =}
( 8)
−
2
×
5.01
= 320.64 awrt 321
(d)
4
X
2
>
5 8
−
X
(
2
1 2
)(
5
)
0
0.5
X
−
X
+
>
⇒
X
>
So need X = 1 or 4 or probability of a + b
= 0.38
Notes
(a) M1
for any correct expression for E(X) in terms of a (or a, b, c)
A1
for a = 0.25
(b) 1st M1
for attempt at an expression for E(X 2) with at least 3 correct nonzero terms
2nd M1
for a correct expression for Var(X) eg"18a – c + 1" – 0.52 Allow with their value
of a substituted
1st A1 for a correct equation for b (or possibly c) eg"18a – c + 1" – 0.52 = 5.01 Allow
with their value of a substituted
2nd A1 for either b = 0.13 or c = 0.24
3rd A1ft for using c = 1 – 2×"0.25" – 2×"0.13" or b = (1 – 2×"0.25" –"0.24" ) ÷2 to gain
the correct ft answer for their 2nd value
(c) B1
for {E(Y) =} 1
M1
for correct use of Var(aX + b) = a2 Var (X )
A1
for awrt 321
(d) 1st M1 for correct quadratic inequality (may be inside prob statement) or table of values
2nd M1 for an attempt to solve or identifying correct X values
1st A1 for X > 0.5 [ may also have X < – 2.5 ]
3rd M1 for realising need X = 1 and 4 only or answer of their (a + b)
2nd A1 for 0.38 (or exact equivalent) only (correct ans only 5/5)
2
{
}
23.7

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