IB DP Math AA Topic: SL 4.12 Standardization of normal variables SL Paper 2

Detail Solution

(a)

(a) For a normal distribution 

, we have:
– Mean 

,
– Standard deviation 

.

First, determine the value of 

 that is 1.5 standard deviations above the mean:
– 1.5 standard deviations = 

,
– Value above the mean: 

.

We need to find 

.

To compute this, standardize 

 into a standard normal variable 

 using the z-score formula:

For 

:

$$Z=\frac{13–10}{2}=\frac{3}{2}=1.5$$

So, 

$P(X>13)=P(Z>1.5)$

.

Using the standard normal distribution:
– The cumulative probability 

$P(Z<1.5)$

 can be found using standard normal tables. Typically, 

$P(Z<1.5)\approx 0.9332$

 (this is a common value from z-tables),
– Therefore, 

$P(Z>1.5)=1–P(Z<1.5)=1–0.9332=0.0668$

.

Thus, the probability that 

$X$

 is more than 1.5 standard deviations above the mean is:

$$P(X>13)=0.0668$$

(b) We need 

. Substituting the mean and standard deviation:

,
So, we need 

.

Standardize this:

Thus, 

, and we’re given:

We need to find the z-score 

 such that 

. Using a standard normal table:

,

.

Since 0.9 lies between these values, 

 is between 1.28 and 1.29. For precision, we can approximate:
The z-score for 0.9 is commonly accepted as 1.2816 (a more precise value from detailed tables).

So, 

 (to two decimal places, as is standard in such problems unless specified otherwise).

————Markscheme—————–

(a) recognition of X > 13 OR  Z > 1.5 (could be seen in a diagram) 
(P (X> 13)=)0.0668072… 
= 0.0668

(b)  EITHER

equating an appropriate correct normal CDF function to 0.1 or 0.9

$P(X>10+2K)=0.1$

$\mathbf{OR}$

$P(Z<K)=0.9$

$\mathbf{OR}$

$P(X<10-2K)=0.1$

$\textbf{OR}$

$P(Z<-K)=0.1$

OR

recognising need to use inverse normal with 0.1 or 0.9

THEN

$1.28155…$

$K=1.28$