Question

The average amount of money spent for lunch per person in the college cafeteria is $6.01 and the standard deviation is $2.19. Suppose that 42 randomly selected lunch patrons are observed. Assume the distribution of money spent is normal, and round all answers to 4 decimal placeswhere possible.

1. What is the distribution of XX? XX ~ N(,)
2. What is the distribution of ¯xx¯? ¯xx¯ ~ N(,)
3. For a single randomly selected lunch patron, find the probability that this patron's lunch cost is between $5.5832 and $5.9922.
4. For the group of 42 patrons, find the probability that the average lunch cost is between $5.5832 and $5.9922.
5. For part d), is the assumption that the distribution is normal necessary? NoYes

Answer 1

a)

X ~ N(6.01,2.19)

b)

x¯ ~ N(6.01,0.3379)

c)

Here, μ = 6.01, σ = 2.19, x1 = 5.5832 and x2 = 5.9922. We need to compute P(5.5832<= X <= 5.9922). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (5.5832 - 6.01)/2.19 = -0.19
z2 = (5.9922 - 6.01)/2.19 = -0.01

Therefore, we get
P(5.5832 <= X <= 5.9922) = P((5.9922 - 6.01)/2.19) <= z <= (5.9922 - 6.01)/2.19)
= P(-0.19 <= z <= -0.01) = P(z <= -0.01) - P(z <= -0.19)
= 0.496 - 0.4247
= 0.0713

d)

Here, μ = 6.01, σ = 0.3379, x1 = 5.5832 and x2 = 5.9922. We need to compute P(5.5832<= X <= 5.9922). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (5.5832 - 6.01)/0.3379 = -1.26
z2 = (5.9922 - 6.01)/0.3379 = -0.05

Therefore, we get
P(5.5832 <= X <= 5.9922) = P((5.9922 - 6.01)/0.3379) <= z <= (5.9922 - 6.01)/0.3379)
= P(-1.26 <= z <= -0.05) = P(z <= -0.05) - P(z <= -1.26)
= 0.4801 - 0.1038
= 0.3763

e)

yes