Question

For n = 40 and π= 0.6, use the normal distribution to approximate the following probabilities a. X 20 b. X>20 c. Xs 20 d. X <20

Answer 1

Sample size , n =    40
Probability of an event of interest, p =   0.60

Mean = np =    24
Variance = np(1-p) =    9.6000
Standard deviation = √variance =    3.0984

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a)

P(X=20) = P(19.5<X<20.5) [ because of continuity correction)

µ =    24                              
σ =    3.0984                              
we need to calculate probability for ,                                  
19.5   ≤ X ≤    20.5                          
X1 =    19.5   ,   X2 =   20.5                  
                                  
Z1 =   (X1 - µ ) / σ =   -1.452                          
Z2 =   (X2 - µ ) / σ =   -1.130                          
                                  
P (   19.5   < X <    20.5   ) =    P (    -1.452368755   < Z <    -1.130   )
                                  
= P ( Z <    -1.130   ) - P ( Z <   -1.452   ) =    0.1293   -    0.0732   =    0.0561

b)

P(X>20) = P(X>20.5) [ because of continuity correction)

as calculated from part a),

P(X<20.5) = 0.1293

sp, P(X>20.5) = 1-0.1293 = 0.8707

c)

P(X≤20) = P(X < 20.5) because of continuity correction)

as calculated from part a),

P(X<20.5) = 0.1293(answer)

d)

P(X<20) = P(X<19.5) because of continuity correction

as calculated from part a),

P(X<19.5) = 0.0732