Question

STAT15_3:

(Please resolve all sections, each section separately, without using Excel software)

The weight of yellow cheese (in grams) on a family pizza is normally split with a 200g span and a standard 15g deviation.

A. What is the probability that the yellow cheese weight on a randomly selected family pizza will be between 194 grams and 209 grams?
B. What is the bottom quarter of the yellow cheese weight distribution on family pizza?
C. Family pizza consists of: dough, sauce, yellow cheese and a possible addition.
The gravy weight is always 100 grams. The weight of the supplement spreads evenly (continuously) in the range of 20 grams to 50 grams.
The various component weights are independent.
John ordered two family pizzas - one with no extra and one with extra.
What is the expectation and what is the difference between the total weight of the two pizzas (without the dough)?

Answer 1

We have:

µ = 200

σ = 15

The test statistic, z = (x - µ)/σ

A. x = 209

z = (209 - 200)/15

z = 0.6

P(z = 0.6) = 0.7257 [From the z-table]

x = 194

z = (194 - 200)/15

z = -0.4

P(z = -0.4) = 0.3446 [From the z-table]

Required probability = 0.7257 - 0.3446 = 0.3812

B. The bottom quarter means the p-value is 0.75.

The z-score for p-value = 0.75 is 0.67. [From the z-table]

z = (x - µ)/σ

0.67 = (x - 200)/15

x = 210.12

C. For the pizza with no extra:

x = 100

z = (100 - 200)/15

z = -6.67

P(z = 6.67) = 0.0000 [From the z-table]

For the pizza with extra:

x = 100 + 50 = 150

z = (150 - 200)/15

z = -3.33

P(z = -3.33) = 0.0004 [From the z-table]

Difference between the total weight of the two pizzas = 150 - 100 = 50