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 , σ^{2} = 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**

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