Given a normal distribution with = 50 and parts (a) through (d). 5, and given yo

Question

Given a normal distribution with = 50 and parts (a) through (d). 5, and given you select a sample of n-100, complete a. What is the probability thatX is less than 49? (Type an integer or deimal rounded to four decimal places as needed.) b. What is the probability that X is between 49 and 51.5? Pl4geXc 51.5) = (Type an integer or decimal rounded to four decimal places as needed.) c. What is the probability that X is above 50.7? PR > 50.7)- (Type an integer or decimal rounded to four decimal places as needed.) d. There is a 40% chance that X is above what value?

Explanation / Answer

the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean of the sampling distribution ( x ) = 50
standard Deviation ( sd )= 5/ Sqrt ( 100 ) =0.5
sample size (n) = 100
a.
P(X < 49) = (49-50)/5/ Sqrt ( 100 )
= -1/0.5= -2
= P ( Z <-2) From Standard NOrmal Table
= 0.02275
b.
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 49) = (49-50)/5/ Sqrt ( 100 )
= -1/0.5
= -2
= P ( Z <-2) From Standard Normal Table
= 0.02275
P(X < 51.5) = (51.5-50)/5/ Sqrt ( 100 )
= 1.5/0.5 = 3
= P ( Z <3) From Standard Normal Table
= 0.99865
P(49 < X < 51.5) = 0.99865-0.02275 = 0.9759
c.
P(X > 50.7) = (50.7-50)/5/ Sqrt ( 100 )
= 0.7/0.5= 1.4
= P ( Z >1.4) From Standard Normal Table
= 0.0808
d.
P ( Z > x ) = 0.4
Value of z to the cumulative probability of 0.4 from normal table is 0.253347
P( x-u / (s.d) > x - 50/5) = 0.4
That is, ( x - 50/5) = 0.253347
--> x = 0.253347 * 5+50 = 51.2667

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