Answer:
(6.35, 5.34)
(6.35, 0.99)
0.10253
0.3984
Step-by-step explanation:
Given that :
μ = 6.35
σ = 2.31
n = 41
What is the distribution of X?X ~ N(,)
X :
Mean of distribution = μ= 6.35
The variance = σ^2 = 2.31^2 = 5.3361 = 5.34
X ~ N = (6.35, 5.34)
What is the distribution of x¯? x¯ ~ N(,)
The mean = μ = 6.35
The standard deviation of the mean = σ/sqrt(n) = 6.35/sqrt(41) = 0.9917 = 0.99
X ~N = (6.35, 0.99)
find the probability that this patron's lunch cost is between $6.1605 and $6.757.
P(6.1605 < x < 6.757)
Obtain the standardized scores:
Z = (x - μ) / σ ; (6.1605 - 6.35) / 2.31 = - 0.082
P(Z < - 0.082) = 0.46732 (Z probability calculator)
(6.757 - 6.35) / 2.31 = 0.176
P(Z < 0.176) = 0.56985 (Z probability calculator)
P(Z < 0.176) -P(Z < - 0.082)
0.56985 - 0.46732 = 0.10253
For the group of 17 patrons, find the probability that the average lunch cost is between $6.1605 and $6.757.
P(6.1605 < x < 6.757) ; n = 17
Obtain the standardized scores:
Z = (x - μ) / σ/sqrt(n) ; (6.1605 - 6.35) / 2.31/sqrt(17) = - 0.338
P(Z < - 0.338) = 0.36768 (Z probability calculator)
(6.757 - 6.35) / 2.31/sqrt(17) = 0.726
P(Z < 0.726) = 0.76608 (Z probability calculator)
P(Z < 0.726) -P(Z < - 0.338)
0.76608 - 0.36768 = 0.3984
