Question
A random variable X is distributed normally with mean 450 and standard deviation 20.
Find \({\rm{P}}(X \le 475)\) .
Given that \({\rm{P}}(X > a) = 0.27\) , find \(a\).
Answer/Explanation
Markscheme
evidence of attempt to find \({\rm{P}}(X \le 475)\) (M1)
e.g. \({\rm{P}}(Z \le 1.25)\)
\({\rm{P}}(X \le 475) = 0.894\) A1 N2
[2 marks]
evidence of using the complement (M1)
e.g. 0.73, \(1 – p\)
\(z = 0.6128\) (A1)
setting up equation (M1)
e.g. \(\frac{{a – 450}}{{20}} = 0.6128\)
\(a = 462\) A1 N3
[4 marks]
Question
A van can take either Route A or Route B for a particular journey.
If Route A is taken, the journey time may be assumed to be normally distributed with mean 46 minutes and a standard deviation 10 minutes.
If Route B is taken, the journey time may be assumed to be normally distributed with mean \(\mu \) minutes and standard deviation 12 minutes.
For Route A, find the probability that the journey takes more than \(60\) minutes.
For Route B, the probability that the journey takes less than \(60\) minutes is \(0.85\).
Find the value of \(\mu \) .
The van sets out at 06:00 and needs to arrive before 07:00.
(i) Which route should it take?
(ii) Justify your answer.
On five consecutive days the van sets out at 06:00 and takes Route B. Find the probability that
(i) it arrives before 07:00 on all five days;
(ii) it arrives before 07:00 on at least three days.
Answer/Explanation
Markscheme
\(A \sim N(46{\text{, }}{10^2})\) \(B \sim N(\mu {\text{, }}{12^2})\)
\({\rm{P}}(A > 60) = 0.0808\) A2 N2
[2 marks]
correct approach (A1)
e.g. \({\rm{P}}\left( {Z < \frac{{60 – \mu }}{{12}}} \right) = 0.85\) , sketch
\(\frac{{60 – \mu }}{{12}} = 1.036 \ldots \) (A1)
\(\mu = 47.6\) A1 N2
[3 marks]
(i) route A A1 N1
(ii) METHOD 1
\({\rm{P}}(A < 60) = 1 – 0.0808 = 0.9192\) A1
valid reason R1
e.g. probability of A getting there on time is greater than probability of B
\(0.9192 > 0.85\) N2
METHOD 2
\({\rm{P}}(B > 60) = 1 – 0.85 = 0.15\) A1
valid reason R1
e.g. probability of A getting there late is less than probability of B
\(0.0808 < 0.15\) N2
[3 marks]
(i) let X be the number of days when the van arrives before 07:00
\({\rm{P}}(X = 5) = {(0.85)^5}\) (A1)
\( = 0.444\) A1 N2
(ii) METHOD 1
evidence of adding correct probabilities (M1)
e.g. \({\rm{P}}(X \ge 3) = {\rm{P}}(X = 3) + {\rm{P}}(X = 4) + {\rm{P}}(X = 5)\)
correct values \(0.1382 + 0.3915 + 0.4437\) (A1)
\({\rm{P}}(X \ge 3) = 0.973\) A1 N3
METHOD 2
evidence of using the complement (M1)
e.g. \({\rm{P}}(X \ge 3) = 1 – {\rm{P}}(X \le 2)\) , \(1 – p\)
correct values \(1 – 0.02661\) (A1)
\({\rm{P}}(X \ge 3) = 0.973\) A1 N3
[5 marks]
Question
In a large city, the time taken to travel to work is normally distributed with mean \(\mu \) and standard deviation \(\sigma \) . It is found that \(4\% \) of the population take less than 5 minutes to get to work, and \(70\% \) take less than 25 minutes.
Find the value of \(\mu \) and of \(\sigma \) .
Answer/Explanation
Markscheme
correct z-values (A1)(A1)
\( – 1.750686 \ldots \) , \(0.524400 \ldots \)
attempt to set up their equations, must involve z-values, not % (M1)
e.g. one correct equation
two correct equations A1A1
e.g. \(\mu – 1.750686\sigma = 5\) , \(0.5244005 = \frac{{25 – \mu }}{\sigma }\)
attempt to solve their equations (M1)
e.g. substitution, matrices, one correct value
\(\mu = 20.39006 \ldots \) , \(\sigma = 8.790874 \ldots \)
\(\mu = 20.4\) \([20.3{\text{, }}20.4]\), \(\sigma = 8.79\) \([8.79{\text{, }}8.80]\) A1A1 N4
[8 marks]
Question
A random variable \(X\) is normally distributed with \(\mu = 150\) and \(\sigma = 10\) .
Find the interquartile range of \(X\) .
Answer/Explanation
Markscheme
recognizing one quartile probability (may be seen in a sketch) (M1)
eg \({\rm{P}}(X < {Q_3}) = 0.75\) , \(0.25\)
finding standardized value for either quartile (A1)
eg \(z = 0.67448 \ldots \) , \(z = – 0.67448 \ldots \)
attempt to set up equation (must be with \(z\)-values) (M1)
eg \(0.67 = \frac{{{Q_3} – 150}}{{10}}\) , \( – 0.67448 = \frac{{x – 150}}{{10}}\)
one correct quartile
eg \({Q_3} = 156.74 \ldots \) , \({Q_1} = 143.25 \ldots \)
correct working (A1)
eg other correct quartile, \({Q_3} – \mu = 6.744 \ldots \)
valid approach for IQR (seen anywhere) (A1)
eg \({Q_3} – {Q_1}\) , \(2({Q_3} – \mu )\)
IQR \( = 13.5\) A1 N4
[7 marks]
Question
The random variable \(X\) is normally distributed with mean \(20\) and standard deviation \(5\).
Find \({\rm{P}}(X \le 22.9)\) .
Given that \({\rm{P}}(X < k) = 0.55\) , find the value of \(k\) .
Answer/Explanation
Markscheme
evidence of appropriate approach (M1)
eg \(z = \frac{{22.9 – 20}}{5}\)
\(z = 0.58\) (A1)
\({\rm{P}}(X \le 22.9) = 0.719\) A1 N3
[3 marks]
\(z\)-score for \(0.55\) is \(0.12566…\) (A1)
valid approach (must be with \(z\)-values) (M1)
eg using inverse normal, \(0.1257 = \frac{{k – 20}}{5}\)
\(k = 20.6\) A1 N3
[3 marks]
Question
The weights of fish in a lake are normally distributed with a mean of \(760\) g and standard deviation \(\sigma \). It is known that \(78.87\% \) of the fish have weights between \(705\) g and \(815\) g.
(i) Write down the probability that a fish weighs more than \(760\) g.
(ii) Find the probability that a fish weighs less than \(815\) g.
(i) Write down the standardized value for \(815\) g.
(ii) Hence or otherwise, find \(\sigma \).
A fishing contest takes place in the lake. Small fish, called tiddlers, are thrown back into the lake. The maximum weight of a tiddler is \(1.5\) standard deviations below the mean.
Find the maximum weight of a tiddler.
A fish is caught at random. Find the probability that it is a tiddler.
\(25\% \) of the fish in the lake are salmon. \(10\% \) of the salmon are tiddlers. Given that a fish caught at random is a tiddler, find the probability that it is a salmon.
Answer/Explanation
Markscheme
Note: There may be slight differences in answers, depending on which values candidates carry through in subsequent parts. Accept answers that are consistent with their working.
(i) \({\text{P}}(X > 760) = 0.5{\text{ (exact), }}[0.499,{\text{ }}0.500]{\text{ }}\) A1 N1
(ii) evidence of valid approach (M1)
recognising symmetry, \(\frac{{0.7887}}{2},{\text{ }}1 – {\text{P}}(W < 815),{\text{ }}\frac{{21.13}}{2} + 78.87\% \)
correct working (A1)
eg\(\;\;\;\)\(0.5 + 0.39435,{\text{ }}1 – 0.10565,\)
\(0.89435{\text{ (exact)}},{\text{ }}0.894{\text{ }}[0.894,{\text{ }}0.895]\) A1 N2
[4 marks]
(i) \(1.24999\) A1 N1
\(z = 1.25{\text{ }}[1.24,{\text{ }}1.25]\)
(ii) evidence of appropriate approach (M1)
eg\(\;\;\;\)\(\sigma = \frac{{x – \mu }}{{1.25}},{\text{ }}\frac{{815 – 760}}{\sigma }\)
correct substitution (A1)
eg\(\;\;\;\)\(1.25 = \frac{{815 – 760}}{\sigma },{\text{ }}\frac{{815 – 760}}{{1.24999}}\)
\(44.0003\)
\(\sigma = 44.0{\text{ }}[44.0,{\text{ }}44.1]{\text{ (g)}}\) A1 N2
[4 marks]
correct working (A1)
eg\(\;\;\;\)\(760 – 1.5 \times 44\)
\(693.999\)
\(694{\text{ }}[693,{\text{ }}694]{\text{ (g)}}\) A1 N2
[2 marks]
\(0.0668056\)
\({\text{P}}(X < 694) = 0.0668{\text{ }}[0.0668,{\text{ }}0.0669]\) A2 N2
[2 marks]
recognizing conditional probability (seen anywhere) (M1)
eg\(\;\;\;\)\({\text{P}}({\text{A}}|{\text{B}}),{\text{ }}\frac{{0.025}}{{0.0668}}\)
appropriate approach involving conditional probability (M1)
eg\(\;\;\;\)\({\text{P}}(S|T) = \frac{{{\text{P}}(S{\text{ and }}T)}}{{{\text{P}}(T)}}\),
correct working
eg\(\;\;\;\)P (salmon and tiddler) \( = 0.25 \times 0.1,{\text{ }}\frac{{0.25 \times 0.1}}{{0.0668}}\) (A1)
\(0.374220\)
\(0.374{\text{ }}[0.374,{\text{ }}0.375]\) A1 N2
[4 marks]
Total [16 marks]
Question
A company makes containers of yogurt. The volume of yogurt in the containers is normally distributed with a mean of \(260\) ml and standard deviation of \(6\) ml.
A container which contains less than \(250\) ml of yogurt is underfilled.
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.
Answer/Explanation
Markscheme
\(0.0477903\)
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]
Question
The masses of watermelons grown on a farm are normally distributed with a mean of \(10\) kg.
The watermelons are classified as small, medium or large.
A watermelon is small if its mass is less than \(4\) kg. Five percent of the watermelons are classified as small.
Find the standard deviation of the masses of the watermelons.
The following table shows the percentages of small, medium and large watermelons grown on the farm.
A watermelon is large if its mass is greater than \(w\) kg.
Find the value of \(w\).
All the medium and large watermelons are delivered to a grocer.
The grocer selects a watermelon at random from this delivery. Find the probability that it is medium.
All the medium and large watermelons are delivered to a grocer.
The grocer sells all the medium watermelons for $1.75 each, and all the large watermelons for $3.00 each. His costs on this delivery are $300, and his total profit is $150. Find the number of watermelons in the delivery.
Answer/Explanation
Markscheme
finding standardized value for 4 kg (seen anywhere) (A1)
eg\(\;\;\;z = – 1.64485\)
attempt to standardize (M1)
eg\(\;\;\;\sigma = \frac{{x – \mu }}{z},{\text{ }}\frac{{4 – 10}}{\sigma }\)
correct substitution (A1)
eg\(\;\;\; – 1.64 = \frac{{4 – 10}}{\sigma },{\text{ }}\frac{{4 – 10}}{{ – 1.64}}\)
\(\sigma = 3.64774\)
\(\sigma = 3.65\) A1 N2
[4 marks]
valid approach (M1)
eg\(\;\;\;1 – p,{\text{ 0.62, }}\frac{{w – 10}}{{3.65}} = 0.305\)
\(w = 11.1143\)
\(w = 11.1\) A1 N2
[2 marks]
attempt to restrict melon population (M1)
eg\(\;\;\;\)\(95\% \) are delivered, \({\text{P}}({\text{medium}}|{\text{delivered}}),{\text{ }}57 + 38\)
correct probability for medium watermelons (A1)
eg\(\;\;\;\frac{{0.57}}{{0.95}}\)
\(\frac{{57}}{{95}},{\text{ }}0.6,{\text{ }}60\% \) A1 N3
[3 marks]
proportion of large watermelons (seen anywhere) (A1)
eg\(\;\;\;{\text{P(large)}} = 0.4,{\text{ }}40\% \)
correct approach to find total sales (seen anywhere) (A1)
eg\(\;\;\;150 = {\text{sales}} – 300,{\text{ total sales}} = \$ 450\)
correct expression (A1)
eg\(\;\;\;1.75(0.6x) + 3(0.4x),{\text{ }}1.75(0.6) + 3(0.4)\)
evidence of correct working (A1)
eg\(\;\;\;1.75(0.6x) + 3(0.4x) = 450,{\text{ }}2.25x = 450\)
200 watermelons in the delivery A1 N2
Notes: If candidate answers 0.57 in part (c), the FT values are \({\text{P(large)}} = 0.43\) and 197 watermelons. Award FT marks if working shown.
Award N0 for 197.
Question
The mass M of apples in grams is normally distributed with mean μ. The following table shows probabilities for values of M.
The apples are packed in bags of ten.
Any apples with a mass less than 95 g are classified as small.
Write down the value of k.
Show that μ = 106.
Find P(M < 95) .
Find the probability that a bag of apples selected at random contains at most one small apple.
Find the expected number of bags in this crate that contain at most one small apple.
Find the probability that at least 48 bags in this crate contain at most one small apple.
Answer/Explanation
Markscheme
evidence of using \(\sum {{p_i}} = 1\) (M1)
eg k + 0.98 + 0.01 = 1
k = 0.01 A1 N2
[2 marks]
recognizing that 93 and 119 are symmetrical about μ (M1)
eg μ is midpoint of 93 and 119
correct working to find μ A1
\(\frac{{119 + 93}}{2}\)
μ = 106 AG N0
[2 marks]
finding standardized value for 93 or 119 (A1)
eg z = −2.32634, z = 2.32634
correct substitution using their z value (A1)
eg \(\frac{{93 – 106}}{\sigma } = – 2.32634,\,\,\frac{{119 – 106}}{{2.32634}} = \sigma \)
σ = 5.58815 (A1)
0.024508
P(X < 95) = 0.0245 A2 N3
[5 marks]
evidence of recognizing binomial (M1)
eg 10, ananaCpqn−=××and 0.024B(5,,)pnp=
valid approach (M1)
eg P(1),P(0)P(1)XXX≤=+=
0.976285
0.976 A1 N2
[3 marks]
recognizing new binomial probability (M1)
eg B(50, 0.976)
correct substitution (A1)
eg E(X) = 50 (0.976285)
48.81425
48.8 A1 N2
[3 marks]
valid approach (M1)
eg P(X ≥ 48), 1 − P(X ≤ 47)
0.884688
0.885 A1 N2
[2 marks]
Question
A random variable X is distributed normally with mean 450. It is known that \({\rm{P}}(X > a) = 0.27\) .
Represent all this information on the following diagram.
Given that the standard deviation is 20, find a . Give your answer correct to the nearest whole number.
Answer/Explanation
Markscheme
A1A1A1 N3
Note: Award A1 for 450 , A1 for a to the right of the mean, A1 for area 0.27 .
[3 marks]
valid approach M1
e.g. \({\rm{P}}(X < a) = 1 – {\rm{P}}(X > a)\) , 0.73
\(a = 462.256 \ldots \) A1
\(a = 462\) A1 N3
[3 marks]