S1 normal help

hiya this is from soloman paper I
A call-centre dealing with complaints collected data on how long customers had to wait
before an operator was free to take their call.
The lower quartile of the data was 12.7 minutes and the interquartile range was 5.8 minutes.
(a) Find the value of the upper quartile of the data. (1 mark) 18.5 mins
It is suggested that a normal distribution could be used to model the waiting time.
(b) Calculate correct to 3 significant figures the mean and variance of this normal distribution based on the values of the quartiles .
(8 marks)
this is what i cant understand....here is the mark scheme
P(Z<12.7 -u/O) = 0.25
12.7-u/O = -0.67
but where does that come from, ive looked in the tables!!
P.s ive used O for variance lol

Reply 1

Newton

a) I. Q. R.=Q3-Q1

5.8=Q3-12.7

=>Q3=18.5 min

b) X~N(m, (s^2))

P(X<Q1)=P(X<12.7)=0.25

P(X<Q3)=P(X<18.5)=0.75

* Z=((X-m)/s)

12.7=>Z=((12.7-m)/s)

Although since we are told that P(X<12.7)=0.25 we know that this value lies to the left of the mean of the distribution, since all values to the left of the mean take negative values we take -Z

P(X<12.7)=P(Z<((m-12.7)/s))=0.25

=>P(X<18.5)=P(Z<((18.5-m)/s))=0.75

Since there is no value of z in the distribution table for which phi(z)=0.25, we say that

Now by applying symmetry

P(Z>((12.7-m)/s))=1-P(Z<((12.7-m)/s))=0.25

=>P(Z<((12.7-m)/s))=0.75

According to the table

=>((12.7-m)/s)=0.67

But from before as the 12.7 lies to the left of the mean,

((m-12.7)/s)=0.67

Which is identical to

((12.7-m)/s)=-0.67

Now to solve the obtained simultaneous equations

18.5-m=0.67s (i)

12.7-m=-0.67s (ii)

(i)+(ii)=>31.2=2m=>m=15.6

=>s=4.3

=>X~N(15.6, (4.3^2))

Newton.