Question
A company produces bags of sugar whose masses, in grams, can be modelled by a normal distribution with mean 1000 and standard deviation 3.5. A bag of sugar is rejected for sale if its mass is less than 995 grams.
Find the probability that a bag selected at random is rejected. [2]
Estimate the number of bags which will be rejected from a random sample of 100 bags. [1]
Given that a bag is not rejected, find the probability that it has a mass greater than 1005 grams. [3]
▶️Answer/Explanation
Ans:
Let \(X\) = mass of a bag of sugar
(a)
evidence of identifying the correct area
P(X<995)= 0.0765637..
= 0.0766
(b) 0.0766 × 100 » 8
(c)
recognition that P(X>1005|X995) is required
\(\frac{P(X ≥ 995\cap X>1005)}{P(X≥ 995)}\)
\(\frac{P(X>1005)}{P(X\geq 995)}\)
\(\frac{0.07656..}{1-0.07656..} (=\frac{0.07656.}{0.9234})\)
=0.0829
Question
After being sprayed with a weedkiller, the survival time of weeds in a field is normally distributed with a mean of 15 days.
(a) If the probability of survival after 21 days is 0.2 , find the standard deviation of the survival time.
When another field is sprayed, the survival time of weeds is normally distributed with a mean of 18 days.
(b) If the standard deviation of the survival time is unchanged, find the probability of survival after 21 days.
▶️Answer/Explanation
Markscheme
(a) required to solve \({\text{P}}\left( {Z < \frac{{21 – 15}}{\sigma }} \right) = 0.8\) (M1)
\(\frac{6}{\sigma } = 0.842 \ldots \,\,\,\,\,\)(or equivalent) (M1)
\( \Rightarrow \sigma = 7.13\) (days) A1 N1
(b) P(survival after 21 days) = 0.337 (M1)A1
[5 marks]
Examiners report
A straightforward Normal distribution problem, but many candidates confused the z value with the probability.
Question
The weight loss, in kilograms, of people using the slimming regime SLIM3M for a period of three months is modelled by a random variable X. Experimental data showed that 67 % of the individuals using SLIM3M lost up to five kilograms and 12.4 % lost at least seven kilograms. Assuming that X follows a normal distribution, find the expected weight loss of a person who follows the SLIM3M regime for three months.
▶️Answer/Explanation
Markscheme
\(X \sim {\text{N}}(\mu ,{\text{ }}{\sigma ^2})\)
\({\text{P}}(X \leqslant 5) = 0.670 \Leftrightarrow \frac{{5 – \mu }}{\sigma } = 0.4399 \ldots \) M1A1
\({\text{P}}(X > 7) = 0.124 \Leftrightarrow \frac{{7 – \mu }}{\sigma } = 1.155 \ldots \) A1
solve simultaneously
\(\mu + 0.4399\sigma = 5{\text{ and }}\mu + 1.1552\sigma = 7\) M1
\(\mu = 3.77{\text{ (3 sf)}}\) A1 N3
the expected weight loss is 3.77 kg
Note: Award A0 for \(\mu = 3.78\) (answer obtained due to early rounding).
[5 marks]
Examiners report
Although many candidates were successful in answering this question, a surprising number of candidates did not even attempt it. The main difficulty was in finding the correct z scores. A fairly common error was to misinterpret one of the conditions and obtain one of the equations as \(\frac{{7 – \mu }}{\sigma } = – 1.155 \ldots \). In some cases candidates failed to keep the accuracy throughout the question and obtained inaccurate answers.
Question
In a factory producing glasses, the weights of glasses are known to have a mean of 160 grams. It is also known that the interquartile range of the weights of glasses is 28 grams. Assuming the weights of glasses to be normally distributed, find the standard deviation of the weights of glasses.
▶️Answer/Explanation
Markscheme
weight of glass = X
\(X \sim {\text{N}}(160,{\text{ }}{\sigma ^2})\)
\({\text{P}}(X < 160 + 14) = {\text{P}}(X < 174) = 0.75\) (M1)(A1)
Note: \({\text{P}}(X < 160 – 14) = {\text{P}}(X < 146) = 0.25\) can also be used.
\({\text{P}}\left( {Z < \frac{{14}}{\sigma }} \right) = 0.75\) (M1)
\(\frac{{14}}{\sigma } = 0.6745 \ldots \) (M1)(A1)
\(\sigma = 20.8\) A1
[6 marks]
Examiners report
Of those students able to start the question, there were good solutions seen. Most students could have made better use of the GDC on this question.
Question
Tim goes to a popular restaurant that does not take any reservations for tables. It has been determined that the waiting times for a table are normally distributed with a mean of \(18\) minutes and standard deviation of \(4\) minutes.
(a) Tim says he will leave if he is not seated at a table within \(25\) minutes of arriving at the restaurant. Find the probability that Tim will leave without being seated.
(b) Tim has been waiting for \(15\) minutes. Find the probability that he will be seated within the next five minutes.
▶️Answer/Explanation
Markscheme
the waiting time, \(X{\text{ ~ }}N\)(\(18\), \({4^2}\))
(a) \({\text{P}}(X > 25) = 0.0401\) (M1)A1
(b) \({\text{P}}(X < 20|X > 15)\) (A1)
\( = \frac{{{\text{P}}(15 < X < 20)}}{{{\text{P}}(X > 15)}}\) (A1)
Note: Only one of the above A1 marks can be implied.
\( = \frac{{0.4648…}}{{0.7733…}} = 0.601\) (M1)A1
[6 marks]
Examiners report
Part (a) was well answered, whilst few candidates managed to correctly use conditional probability for part (b).
Question
Bob measured the heights of 63 students. After analysis, he conjectured that the height, \(H\) , of the students could be modelled by a normal distribution with mean 166.5 cm and standard deviation 5 cm.
(a) Based on this assumption, estimate the number of these students whose height is at least 170 cm.
Later Bob noticed that the tape he had used to measure the heights was faulty as it started at the 5 cm mark and not at the zero mark.
(b) What are the correct values of the mean and variance of the distribution of the heights of these students?
▶️Answer/Explanation
Markscheme
(a) \(H{\text{ \~ N}}\)(\(166.5\), \({5^2}\) )
\({\text{P}}(H \geqslant 170) = 0.242…\) (M1)(A1)
\(0.242… \times 63 = 15.2\) A1
so, approximately \(15\) students
(b) correct mean: \(161.5\) (cm) A1
variance remains the same, i.e. 25 (cm2) A2
[6 marks]
Examiners report
A surprising number of students lacked the basic knowledge of the normal distribution and were unable to answer the first part of this question. Those students who showed a knowledge of the topic tended to answer the question well. In part (b) many students either had a misunderstanding of the difference between variance and standard deviation, or did not read the question properly.
Question
A company produces computer microchips, which have a life expectancy that follows a normal distribution with a mean of 90 months and a standard deviation of 3.7 months.
(a) If a microchip is guaranteed for 84 months find the probability that it will fail before the guarantee ends.
(b) The probability that a microchip does not fail before the end of the guarantee is required to be 99 %. For how many months should it be guaranteed?
(c) A rival company produces microchips where the probability that they will fail after 84 months is 0.88. Given that the life expectancy also follows a normal distribution with standard deviation 3.7 months, find the mean.
▶️Answer/Explanation
Markscheme
(a) \({\text{P}}(X \leqslant 84) = {\text{P}}(Z \leqslant – 1.62…) = 0.0524\) (M1)A1 N2
Note: Accept 0.0526.
(b) \({\text{P}}(Z \leqslant z) = 0.01 \Rightarrow z = – 2.326…\) (M1)
\({\text{P}}(X \leqslant x) = {\text{P}}(Z \leqslant z) = 0.01 \Rightarrow z = – 2.326…\)
\(x = 81.4\,\,\,\,\,{\text{(accept 81)}}\) A1 N2
(c) \({\text{P}}(X \leqslant 84) = 0.12 \Rightarrow z = – 1.1749…\) (M1)
\({\text{mean is 88.3}}\,\,\,\,\,{\text{(accept 88)}}\) A1 N2
[6 marks]
Examiners report
A fair amount of students did not use their GDC directly, but used tables and more traditional methods to answer this question. Part (a) was answered correctly by most candidates using any method. A large number of candidates reversed the probabilities, i.e., failed to use a negative z value in parts (b) and (c), and hence did not obtain correct answers.
Question
In a factory producing glasses, the weights of glasses are known to have a mean of 160 grams. It is also known that the interquartile range of the weights of glasses is 28 grams. Assuming the weights of glasses to be normally distributed, find the standard deviation of the weights of glasses.
▶️Answer/Explanation
Markscheme
weight of glass = X
\(X \sim {\text{N}}(160,{\text{ }}{\sigma ^2})\)
\({\text{P}}(X < 160 + 14) = {\text{P}}(X < 174) = 0.75\) (M1)(A1)
Note: \({\text{P}}(X < 160 – 14) = {\text{P}}(X < 146) = 0.25\) can also be used.
\({\text{P}}\left( {Z < \frac{{14}}{\sigma }} \right) = 0.75\) (M1)
\(\frac{{14}}{\sigma } = 0.6745 \ldots \) (M1)(A1)
\(\sigma = 20.8\) A1
[6 marks]
Examiners report
Of those students able to start the question, there were good solutions seen. Most students could have made better use of the GDC on this question.