e amount of paint required to paint a surface with an area of 50 m2 is normally
ID: 3323691 • Letter: E
Question
e amount of paint required to paint a surface with an area of 50 m2 is normally distributed with a mean 6 Liter and a standard deviation of 0.3 Liter. (1) If 6.5 Liter paint are available, what is the probability that entire surface can be painted? A. 0.95 8.0.85 C. 1.0 D. 0.76 E.0.5 (2) Find the I1QR (inter quartile range) of the distribution. A. 0.6 B.0.75 c. 0.4 D. 0.9 E.0.55 (3) What must the standard deviation be so that the probability is 0.9 that 6.2 L paint will be sufficient to paint the entire surface? A. 0.25 B.0.3 C. 0.4D.0.63 E.0.16 (4) What is the probability you can complete the task with 5 L of paint? A. 0.01 B. 0.03c.000D. 0.1 E. 0.06Explanation / 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 ~ N(0,1)
mean ( u ) = 6
standard Deviation ( sd )= 0.3
Q1.
P(X < 6.5) = (6.5-6)/0.3
= 0.5/0.3= 1.6667
= P ( Z <1.6667) From Standard Normal Table
= 0.9522 ~ 0.95
Q2.
P ( Z < x ) = 0.25
Value of z to the cumulative probability of 0.25 from normal table is -0.67449
P( x-u/s.d < x - 6/0.3 ) = 0.25
That is, ( x - 6/0.3 ) = -0.67449
--> x = -0.67449 * 0.3 + 6 = 5.797653
P ( Z < x ) = 0.75
Value of z to the cumulative probability of 0.75 from normal table is 0.67449
P( x-u/s.d < x - 6/0.3 ) = 0.75
That is, ( x - 6/0.3 ) = 0.67449
--> x = 0.67449 * 0.3 + 6 = 6.202347
Inter quartile = Q3 - Q1 = 6.202347 - 5.797653 = 0.404694 ~ 0.40
Q3.
P ( Z < X ) = 0.9
Value of z to the cumulative probability of 0.9 from normal table is 1.281552
P( X-u/s.d < X - 6/ s.d ) = 0.9
That is, ( 6.2 - 6/ s.d ) = 1.281552
--> s.d = ( 6.2 - 6 ) / 1.281552
--> s.d = ( 0.2 ) / 1.281552
--> s.d = 0.16
Q4.
P(X < 5) = (5-6)/0.3
= -1/0.3= -3.3333
= P ( Z <-3.3333) From Standard Normal Table
= 0.0004 ~ 0.00
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