Question

The average amount of money spent for lunch per person in the college cafeteria is $6.93 and the standard deviation is $2.61. Suppose that 17 randomly selected lunch patrons are observed. Assume the distribution of money spent is normal, and round all answers to 4 decimal placeswhere possible.

What is the distribution of X?X ~ N(,)

What is the distribution of x¯? x¯ ~ N(,)

For a single randomly selected lunch patron, find the probability that this patron's lunch cost is between $6.1605 and $6.757.

For the group of 17 patrons, find the probability that the average lunch cost is between $6.1605 and $6.757.

Answer 1

Part a)
X ~ N ( µ = 6.93 , σ² = 6.8121 )

Part b)
µ_X̅ = µ = 6.93
σ_X̅ = σ / √ (n) = 2.61/√17 = 0.633

X ~ N ( µ = 6.93 , σ = 0.633 )

Part c)
X ~ N ( µ = 6.93 , σ = 2.61 )
P ( 6.1605 < X < 6.757 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 6.1605 - 6.93 ) / 2.61
Z = -0.2948
Z = ( 6.757 - 6.93 ) / 2.61
Z = -0.0663
P ( -0.29 < Z < -0.07 )
P ( 6.1605 < X < 6.757 ) = P ( Z < -0.07 ) - P ( Z < -0.29 )
P ( 6.1605 < X < 6.757 ) = 0.4736 - 0.3841
P ( 6.1605 < X < 6.757 ) = 0.0895

Part d)
X ~ N ( µ = 6.93 , σ = 2.61 )
P ( 6.1605 < X < 6.757 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 6.1605 - 6.93 ) / ( 2.61 / √(17))
Z = -1.2156
Z = ( 6.757 - 6.93 ) / ( 2.61 / √(17))
Z = -0.2733
P ( -1.22 < Z < -0.27 )
P ( 6.1605 < X̅ < 6.757 ) = P ( Z < -0.27 ) - P ( Z < -1.22 )
P ( 6.1605 < X̅ < 6.757 ) = 0.3923 - 0.1121
P ( 6.1605 < X̅ < 6.757 ) = 0.2802