Question

Coffee shop customers are charged for salad they takefrom the salad bar. Sampling suggests that the amount of salad taken is uniformly distributed between 160g and 450g.
1) Draw a graph of this distribution.
2) Show that the area of this uniform distribution is 1.00.
3) What is the mean of salad plate filling Weight? What is the standard deviation of salad plate filling Weight?
4) What is the probability customer will take more than 360 grams?
5)What is the probability that a customer will take between 336 and 420 grams of salad?

Answer 1

Solution :-

given that, salad is taken uniformly distributed between 160g and 450g.

so,

• smallest value of salad taken = a = 160 .
• smallest value of salad taken = b = 450 .

then,

→ f(x) = 1/(b - a) for a ≤ x ≤ b

→ f(x) = 1/(450 - 160)

→ f(x) = (1/290)

now,

→ The mean of salad plate filling Weight = (a + b)/2 = (160 + 450)/2 = 610/2 = 305 g .

and,

→ Standard deviation of salad plate filling Weight = √[(b - a)²/12] = √[(450 - 290)²/12] = √[(290)²/12] = √(7008.34) ≈ 83.72 g .

now,

→ The probability customer will take more than 360 grams = 360 < x < 450 => f(360 < x < 450) = (450 - 360) * f(x) = 90 * (1/290) = (9/29) .

and,

→ The probability that a customer will take between 336 and 420 grams of salad = 336 ≤ x ≤ 420 => f(336 ≤ x ≤ 420) = (420 - 336) * f(x) = 84 * (1/290) = (42/145) .

Learn more :-

The 99% confidence interval for mean is (12, 16).

Then, find the values of sample mean

and margin of error.

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