Question

Given a normal distribution with p= 104 and o = 20, and given you select a sample of n= 16, complete parts (a) through (d). Click here to view page 1 of the cumulative standardized normal distribution table. Click here to view page 2 of the cumulative standardized normal distribution table. a. What is the probability that X is less than 94? P(<94) = 0.0228 (Type an integer or decimal rounded to four decimal places as needed.) b. What is the probability that X is between 94 and 96.5? P(94 <x< 96.5) =] (Type an integer or decimal rounded to four decimal places as needed.) c. What is the probability that X is above 105.4? P(X> 105.4) = 0 (Type an integer or decimal rounded to four decimal places as needed.) d. There is a 63% chance that X is above what value? X=1 (Type an integer or decimal rounded to two decimal places as needed.)

Answer 1

(a)

z-score for $\bar{x}=94$ is

$z=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\frac{94-104}{20/\sqrt{16}}=-2$

So

$P(\bar{x}<94)=P(z<-2)=0.0228$

(b)

z-score for $\bar{x}=96.5$ is

$z=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\frac{96.5-104}{20/\sqrt{16}}=-1.5$

So

$P(94<\bar{x}<96.5)=P(-2<z<-1.5)=P(z\leq -1.5)-P(z\leq -2)=0.0668-0.0228=0.0440$

(c)

z-score for $\bar{x}=105.4$ is

$z=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\frac{105.4-104}{20/\sqrt{16}}=0.28$

So  

$P(\bar{x}>105.4)=P(z>0.28)=1-P(z\leq 0.28)=1-0.6103=0.3897$

(d)

Here we need z-score that has 0.63 area to its right. z-score -0.33 has 0.63 area to its right and 0.37 area to its left. So required sample mean is

$-0.33=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\frac{\bar{x}-104}{20/\sqrt{16}}$

$\bar{x}=102.35$