Independent probability calculations

5 The lengths, in cm, of the leaves of a particular type are modelled by the distribution \(\mathrm { N } \left( 5.2,1.5 ^ { 2 } \right)\).

1. Find the probability that a randomly chosen leaf of this type has length less than 6 cm .
   The lengths of the leaves of another type are also modelled by a normal distribution. A scientist measures the lengths of a random sample of 500 leaves of this type and finds that 46 are less than 3 cm long and 95 are more than 8 cm long.
2. Find estimates for the mean and standard deviation of the lengths of leaves of this type.
3. In a random sample of 2000 leaves of this second type, how many would the scientist expect to find with lengths more than 1 standard deviation from the mean?

2 The lengths of the rods produced by a company are normally distributed with mean 55.6 mm and standard deviation 1.2 mm .

1. In a random sample of 400 of these rods, how many would you expect to have length less than 54.8 mm ?
2. Find the probability that a randomly chosen rod produced by this company has a length that is within half a standard deviation of the mean.

4 A \(\boldsymbol { \rho }\) trb station fid th tits \(\mathbf { d }\) ily sales，in litres，are \(\mathbf { n }\) mally \(\dot { \mathbf { d } }\) strib ed with mean ad stad rd d \(\dot { \mathbf { v } }\) atin \(\quad 0\)
（a）Fid 0 may dy 6 th \(\mathbf { y }\) ar（B d \(\mathbf { y }\) ）th d ily sales can b eq cted to e区 eed \(\boldsymbol { \theta }\) litres． Th d ily sales at an \(\mathbf { b } r \mathbf { p }\) trb station are \(X\) litres，we re \(X\) is \(\mathbf { n }\) mally \(\dot { \mathbf { d } }\) strib ed with mean \(m\) ad stad rd iv atird \(\quad t\) is g it h \(\mathrm { t } \mathrm { P } ( X > 0 = \mathbb { 0 }\)
（b）Fid by le \(6 m\) ．
(c) Fid th p b b lity th t d ily sales at th s p trb station ex eed \(\theta\) litres \(\mathbf { n }\) fewer th n 266 rach lyc \(b\) end \(y\).
[0pt] [ \(\beta\)

5 A general store sells lawn fertiliser in 2.5 kg bags, 5 kg bags and 10 kg bags.

1. The actual weight, \(W\) kilograms, of fertiliser in a 2.5 kg bag may be modelled by a normal random variable with mean 2.75 and standard deviation 0.15 . Determine the probability that the weight of fertiliser in a 2.5 kg bag is:
   1. less than 2.8 kg ;
   2. more than 2.5 kg .
2. The actual weight, \(X\) kilograms, of fertiliser in a 5 kg bag may be modelled by a normal random variable with mean 5.25 and standard deviation 0.20 .
   1. Show that \(\mathrm { P } ( 5.1 < X < 5.3 ) = 0.372\), correct to three decimal places.
   2. A random sample of four 5 kg bags is selected. Calculate the probability that none of the four bags contains between 5.1 kg and 5.3 kg of fertiliser.
3. The actual weight, \(Y\) kilograms, of fertiliser in a 10 kg bag may be modelled by a normal random variable with mean 10.75 and standard deviation 0.50. A random sample of six 10 kg bags is selected. Calculate the probability that the mean weight of fertiliser in the six bags is less than 10.5 kg .

3. A study was made of the heights of boys of different ages in Lancashire. The study concluded that the heights of 13 year-old boys are normally distributed with a mean of 156 cm and a variance of \(73 \mathrm {~cm} ^ { 2 }\). Find the probability that a 13 year-old boy chosen at random will be

1. more than 165 cm tall,
2. between 156 and 165 cm tall. The study also concluded that the heights of 14 year-old boys are normally distributed with a mean of 160 cm and a variance of \(79 \mathrm {~cm} ^ { 2 }\). One 13 year-old and one 14 year-old boy are chosen at random.
3. Find the probability that both boys are more than 165 cm tall.
4. State, with a reason, whether the probability that the combined height of the two boys is more than 330 cm is more or less than your answer to part (c).
   (2 marks)

6 When Monica walks to work from home, she uses either route A or route B.

1. Her journey time, \(X\) minutes, by route A may be assumed to be normally distributed with a mean of 37 and a standard deviation of 8 . Determine:
   1. \(\mathrm { P } ( X < 45 )\);
   2. \(\mathrm { P } ( 30 < X < 45 )\).
2. Her journey time, \(Y\) minutes, by route B may be assumed to be normally distributed with a mean of 40 and a standard deviation of \(\sigma\). Given that \(\mathrm { P } ( Y > 45 ) = 0.12\), calculate the value of \(\sigma\).
3. If Monica leaves home at 8.15 am to walk to work hoping to arrive by 9.00 am , state, with a reason, which route she should take.
4. When Monica travels to work from home by car, her journey time, \(W\) minutes, has a mean of 18 and a standard deviation of 12 . Estimate the probability that, for a random sample of 36 journeys to work from home by car, Monica's mean time is more than 20 minutes.
5. Indicate where, if anywhere, in this question you needed to make use of the Central Limit Theorem.

5

1. Wooden lawn edging is supplied in 1.8 m length rolls. The actual length, \(X\) metres, of a roll may be modelled by a normal distribution with mean 1.81 and standard deviation 0.08 . Determine the probability that a randomly selected roll has length:
   1. less than 1.90 m ;
   2. greater than 1.85 m ;
   3. between 1.81 m and 1.85 m .
2. Plastic lawn edging is supplied in 9 m length rolls. The actual length, \(Y\) metres, of a roll may be modelled by a normal distribution with mean \(\mu\) and standard deviation \(\sigma\). An analysis of a batch of rolls, selected at random, showed that $$\mathrm { P } ( Y < 9.25 ) = 0.88$$
   1. Use this probability to find the value of \(z\) such that $$9.25 - \mu = z \times \sigma$$ where \(z\) is a value of \(Z \sim \mathrm {~N} ( 0,1 )\).
   2. Given also that $$\mathrm { P } ( Y > 8.75 ) = 0.975$$ find values for \(\mu\) and \(\sigma\).