Answer to Question #139631 in Statistics and Probability for dj

Question

3. Consider a normal distribution with a mean of 500 and a standard deviation of 50.
a. Below what value can we expect to have the lowest 20%?
b. Between what values can we expect to find the middle 80%?

Answer 1

$\mu =500, \sigma=50$

a) $P(X<N)=20\%=0.2$

Z score is -0.84 then

$\frac{N-500}{50}=-0.84 \implies N=500-50*0.84=458$

b) $P(N_1<X<N_2)=80\%$

$P(X<N_2)=90\%=0.9\implies$ Z score is 1.28

$\frac{N_2-500}{50}=1.28 \implies N_2=500+50*1.28=564$

$P(X<N_1)=10\%=0.1\implies$ Z score is -1.28

$\frac{N_1-500}{50}=-1.28 \implies N_1=500-50*1.28=436$

Answer:

a) 458

b) between 436 and 564