Question

This Test: 30 pls possible The percentage of the members of a certain population who have weekly Incomes between 597 51 and 5154 06h 40%. It is known that weekly incomes for this population are approximately normally Gributed with a mean af 12 one member of this population is selected at random, what is the probability he or she will have a weekly income that is less than 5154 067 OA 027 OB 023 OC. 0.45 OD. 0.73 OE. None of the above Type here to search ( 5 4 7 6 3 9 8 R T Y U P.

Answer 1

Let X = weekly income of the randomly selected person.

Then X~ Normal(126 , $\sigma^2$ ) $\Rightarrow$ (X - 126) / $\sigma$ ~ N(0,1). Hence we can compute the probabilities using the Standard Normal Tables:
Let Z = (X - 126) / $\sigma$ $\Rightarrow$ Z~N(0,1)

To find $\sigma$ we'll use the infromation that P(97.94 <= X <= 154.06) = 0.46.

Now, P(97.94 <= X <= 154.06) = P( (97.94 - 126) / $\sigma$ <= (X - 126) / $\sigma$ <= (154.06 - 126) / $\sigma$ )
= P( (97.94 - 126) / $\sigma$ <= Z <= (154.06 - 126) / $\sigma$ ) = P( -28.06/$\sigma$ <= Z <= 28.06/$\sigma$ )
=P( Z <= 28.06/$\sigma$ ) - P(Z <= -28.06/$\sigma$ )
=2*P(Z<= 28.06/$\sigma$ ) - 1 [ from the symmetricity of N(0,1) : P( Z <= 28.06/$\sigma$ ) - P(Z <= -28.06/$\sigma$ )=1 ]
Now,
P(97.94 <= X <= 154.06) = 0.46.
$\Rightarrow$ 2*P(Z<= 28.06/$\sigma$ ) - 1 = 0.46.
$\Rightarrow$ P(Z<= 28.06/$\sigma$ ) = 0.73
From N(0,1) standard normal Tables it is verified that P(Z <= 0.61281) = 0.73
$\Rightarrow$ 28.06/$\sigma$ = 0.61281 $\Rightarrow$   $\sigma$ = 45.789

Therefore, X ~ Normal(126 , 45.789² )
Now we can answer the question of P( X<=  154.06) =  0.7299 = 0.723 (answer)