Question

Q9 Using the standard normal distribution table, answer the following questions:

In a study related to pregnant women, it was observed that the hemoglobin values of this group showed normal distribution and the mean was 12.5 gr / dl and the standard deviation was 1.0 gr / dl. Accordingly, pregnant women; What is the probability of the hemoglobin values being between 11.5‐13.0 gr / dl? What symmetrical limits are the hemoglobin values of 80% of? What are the hemoglobin values of 30% of the top? What is the lower hemoglobin values of 22% of the lower? What percentage of hemoglobin value is less than 12.2 gr / dl?

Answer 1

a)

probability of the hemoglobin values being between 11.5‐13.0 gr / dl:

probability =P(11.5<X<13)=P((11.5-12.5)/1)<Z<(13-12.5)/1)=P(-1<Z<0.5)=0.6915-0.1587=0.5328

b)

for middle 80% values ; critical z =1.28

corresponding middle 80% inteval =(12.5-1.28*1 ; 12.5+1.28*1) = (11.22 to 13.78)

c)

for top 0 percent or 70th percentile critical value of z=		0.52
therefore corresponding value=mean+z*std deviation=12.5+0.52*1 =			13.02

d)

for 22th percentile critical value of z=		-0.77
therefore corresponding value=mean+z*std deviation=12.5-0.77*1 =			11.73

e)

What percentage of hemoglobin value is less than 12.2 gr / dl:

probability =P(X<12.2)=(Z<(12.2-12.5)/1)=P(Z<-0.3)=0.3821