Given a normal distribution with u 70 and standard deviation

Given a normal distribution with u= 70 and standard deviation =7, if you select a sample of n=81, what is the probability that the sample mean is

a. less than 68.5?

b. between 68.5 and 69.5?

c. above 71.6?

d. there is a 35% chance that sample mean (x bar) above what value?

Given a normal distribution with u= 70 and standard deviation =7, if you select a sample of n=81, what is the probability that the sample mean is

Standard error = sd/sqrt(n)=7/sqrt(81) =0.7778

z value of 68.5, z=(68.5-70)/0.7778 =-1.93

P( mean x <68.5) = p( z < -1.93) = 0.0268

z value of 69.5, z=(69.5-70)/0.7778 =-0.64

P(68.5 < mean x < 69.5)

P( -1.93 <z<-0.64)

=P( z< -0.64)-P( z <-1.93)

= 0.2611 - 0.0268

=0.2343

z value of 71.6, z=(71.6-70)/0.7778 =2.06

P( mean x >71.6)=P(z > 2.06) = 0.0197

d. there is a 35% chance that sample mean (x bar) above what value?

Z value for top 35% =0.385

(x bar) = mean+z*se = 70+0.385*0.7778 = 70.299453

The required value =70.30 ( two decimals).