Question

The average amount of money spent for lunch per person in the college cafeteria is $6.35 and the standard deviation is $2.31. Suppose that 41 randomly selected lunch patrons are observed. Assume the distribution of money spent is normal, and round all answers to 4 decimal places where 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

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