Let X be normally distributed with mean = 4.3 and standard deviation = 2. Use Ta

Question

Let X be normally distributed with mean = 4.3 and standard deviation = 2. Use Table 1.

a. Find P(X > 6.5). (Round "z" value to 2 decimal places and final answer to 4 decimal places.)

           P(X > 6.5)_______________

b. Find P(5.5 X 7.5). (Round "z" value to 2 decimal places and final answer to 4 decimal places.)

              P(5.5 X 7.5)_______________

c. Find x such that P(X > x) = 0.0869. (Round "z" value and final answer to 3 decimal places.)

                x ___________________

d. Find x such that P(x X 4.3) = 0.2088. (Negative value should be indicated by a minus sign. Round "z" value and final answer to 3 decimal places.)

          x_________________

Explanation / Answer

We have = 4.3 and standard deviation = 2, i.e., X~ N(4.3, 2)

a) P(X > 6.5) = ?

When X= 6.5, Z= (6.5-4.3)/2 = 1.1

Therefore,  P(X > 6.5) =  P(X >1.1) = 0.1357

b) P(5.5 X 7.5) = ?

When X= 5.5, Z= (5.5-4.3)/2 = 0.6

When X= 7.5, Z= (7.5-4.3)/2 = 1.6

Therefore,  P(5.5 X 7.5) =   P(0.6 X 1.6) = P(X 1.6)-P( X 0.6)=0.9452-0.7257 =0.219

c) P(x X 4.3) = 0.2088 (1)

When X= x, Z= (x-4.3)/2

When X= 4.3, Z= (4.3-4.3)/2 = 0

Therefore, P(x X 4.3)=   P( (x-4.3)/2 X 0)  = P(X 0)-P( X (x-4.3)/2)=0.5 - P( X   (x-4.3)/2)

   P(x X 4.3) = 0.5 - P( X   (x-4.3)/2) (2)

From (1) and (2)

0.5 - P( X   (x-4.3)/2) =   0.2088

=> P( X   (x-4.3)/2) = 0.2912

We can calculate the Z value which gave the 0.2912 probability from the Normal table. The Z value which is give the probability 0.2912 is -0.5499.

i.e.,  P( X   -0.5499) = 0.2912

Therefore,  (x-4.3)/2=  -0.5499

=>  (x-4.3) =-1.0932

=> x = -1.0932 + 4.3

=> x= 3.2068

The value of x which give  that P(x X 4.3) = 0.2088 is 3.2068 (3.207 rounded at 3 decimal place).

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