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The chicken
and the egg
Statistics 2 – Activity 4
1
The chicken and the egg
Eggs are often graded by size by weight
Medium
Large
Extra Large
If X is the distribution of the weight of eggs in gms then we can write
X ~ N(58, 52)
If ‘extra large’ eggs weigh more than 66gms, what is the probability
of an egg chosen at random being ‘extra large’?
If X ~ N(58, 52), then we need to calculate P(X>66)
To do this we need to transform our distribution into a
standardised normal distribution.
We do this by changing x values into z values using this
transformation.
What happens when we use this transformation?
Since X ~ N(58,52), μ = 58 and σ = 5
Firstly see what happens to the mean of our X distribution
So the mean is transformed into 0
Now what happens to the standard deviation of our X distribution?
Look at an X value that is 1 standard deviation above the mean
i.e. 58 + 5 = 63
So for the new distribution, the mean + 1 standard deviation = 1
but the mean is 0, so the new standard deviation is 1.
Our new distribution is given by Z ~ N(0,1)
If X ~ N(58, 52), then we need to calculate P(X>66)
This has now been changed into a standardised normal distribution Z ~ N(0, 1).
What about our key number 66?
If x = 66, then z = 66 – 58 = 1.6
5
Our problem can be illustrated by this diagram
And our problem becomes P(Z>1.6) which you should be able to do!
P(Z>1.6)
P(Z>1.6) = 1 – P(Z<1.6) = 1 – Φ(1.6)
P(Z>1.6) = 1 – 0.9452 = 0.0548 or 5.5%
So there is a 5.5% chance of a randomly chosen egg being
‘extra large’.
Back to Jake
In ‘Baby boom’ we thought about quite a special
baby. Jake, baby son of Jasmine and Jeremy
Whitehouse, weighed in at a whopping 10.8Ibs –
the heaviest baby born at the Royal Hospital for
over five years.
We already know
that Jake is pretty
special – but just
how rare are babies
that weigh 10.8Ibs
or above?
When asked, a data analyst from the National Statistics Office
stated that…
So if X is the distribution of all babies born in the UK, then
we can write
X ~ N(7.18, 1.212)
We need to find P(X>10.8)
X ~ N(7.18,1.212)
We need to find P(X>10.8)
To do this we standardise the distribution
z = 10.8 – 7.18 = 2.992
1.21
P(Z>2.992) = 1 – Φ(2.992)
= 1 – 0.9986
= 0.0014 or 0.14%
So the chances of a newborn baby weighing greater than
10.8 lbs is 0.14% – or about one in every 700 babies.
This demonstrates just how special baby Jake actually is!
What about baby Josie?
You may recall that on the same day,
baby Josie Allchurch weighed in at 5.8lbs.
So can we now decide how rare this is?
We need to find P(X<5.8)
X ~ N(7.18,1.212)
We need to find P(X<5.8)
To do this we standardise the distribution
z = 5.8 – 7.18 = –1.1405
1.21
We need to find Φ(–1.1405)
In reality, it is easier to find P(Z>1.1405)
P(Z>1.1405) = 1 - Φ(1.1405)
= 1 - 0.8730
= 0.1270 or 12.7%
So is baby Josie special?
The chances of a newborn
baby weighing less than 5.8lbs
is 12.7%, or about one in every
eight babies. So, Josie is not
that special in terms of her
weight – but she is special to
her Mum and Dad!
Suggestion
Now try ‘Practice time’
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