Question

"Black Friday" is the day after Thanksgiving and the traditional first day of the Christmas shopping season. Suppose a recent poll suggested that 66% of Black Friday shoppers are actually buying for themselves. A random sample of 130 Black Friday shoppers is obtained. Answer each problem using the normal approximation to the binomial distribution.

(a)

Find the approximate probability that fewer than 73 Black Friday shoppers are buying for themselves. (Round your answer to four decimal places.)

(b)

Find the approximate probability that between 74 and 84 (inclusive) Black Friday shoppers are buying for themselves. (Round your answer to four decimal places.)

Answer 1

Mean = n * P = ( 130 * 0.66 ) = 85.8
Variance = n * P * Q = ( 130 * 0.66 * 0.34 ) = 29.172
Standard deviation = √(variance) = √(29.172) = 5.4011

Part a)

P ( X < 73 )
Using continuity correction
P ( X < n - 0.5 ) = P ( X < 73 - 0.5 ) = P ( X < 72.5 )

X ~ N ( µ = 85.8 , σ = 5.4011 )
P ( X < 72.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 72.5 - 85.8 ) / 5.4011
Z = -2.46
P ( ( X - µ ) / σ ) < ( 72.5 - 85.8 ) / 5.4011 )
P ( X < 72.5 ) = P ( Z < -2.46 )
P ( X < 72.5 ) = 0.0069

Part b)

P ( 74 <= X <= 84 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 74 - 0.5 < X < 84 + 0.5 ) = P ( 73.5 < X < 84.5 )

X ~ N ( µ = 85.8 , σ = 5.4011 )
P ( 73.5 < X < 84.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 73.5 - 85.8 ) / 5.4011
Z = -2.28
Z = ( 84.5 - 85.8 ) / 5.4011
Z = -0.24
P ( -2.28 < Z < -0.24 )
P ( 73.5 < X < 84.5 ) = P ( Z < -0.24 ) - P ( Z < -2.28 )
P ( 73.5 < X < 84.5 ) = 0.4052 - 0.0113
P ( 73.5 < X < 84.5 ) = 0.3939