Question

Do the following problems. Begin each problem by restating the question. Neatness counts. 25 points.

Minitab may be used, attach or include your output if you use Minitab to find answers.

1. The weekly demand for Domino’s pizzas on a Friday night is a normally distributed random variable with mean 235 and standard deviation 10, N(235,10). Show your work. (15 points)
   • Find the probability that demand exceeds 220. (2 points)
   • Find the probability that demand is between 230 and 245. (3 points)
   • Find the probability that demand greater than 250. (2 points)
   • Find the value of X for the lowest 25 percent. (4 points)
   • Suppose the store wants to make sure they are able to meet demand 90% of the time, how many pizzas should they bake? (4 points)

2. Shower temperature at the Spokane Club showers is regulated automatically. The heater kicks in when the temperature falls to 98⁰F and shuts off when the temperature reaches 108⁰ Water temperature then falls slowly until the heater kicks in again. At a given moment, the water temperature is a uniformly distributed random variable U(98,108). (10 points)
   • Find the mean temperature. (2 points)
   • Find the standard deviation of the temperature. (2 points)
   • Find the 25th percentile for water temperature. (3 points)
   • Find the probability that the temperature is over 101⁰ (3 points)

Answer 1

Ans 1. The weekly demand for Domino’s pizzas on a Friday night is a normally distributed random variable with mean 235 and standard deviation 10, N(235,10). Show your work. (15 points)

You basically need to put up a normal distribution with the given statistics.

Let the random variable X denote the weekly demand. We are given-

X ~ N(μ, σ) where μ = 235 and σ = 10.

Take a look at the image below:

p(X) μ-235 215 225 245 255

a) Find the probability that demand exceeds 220. (2 points)

To calculate this, we'd have to standardise X. In other words, we'd have to calculate the Z variable corresponding to 220.

Z = (X-μ) / σ = (220-235)/10 = -1.5

Now from the Z-table, let's check the probability of Z > -1.5 (since you need the probability of X > 220)

P(Z>-1.5) = P(-1.5<Z<0) + P(0<Z) = P(0<Z<1.5) + P(0<Z) (By symmetry: Normal distribution curve has the same shape on either side of the line X=μ)

P(Z>-1.5) = 0.4332 + 0.5 = 0.9332

Hence, there's a 93.32% chance that the demand exceeds 220.

b) Find the probability that demand is between 230 and 245. (3 points)

Just like in a) above, we'd calculate the Z values (and their probabilities) of 230 and 245 first.

We get-

Z₂₃₀ = -0.5 and Z₂₄₅ = 1

P (0>Z>-0.5) = 0.1915 and P(0<Z<1) = 0.3413

Adding the two we get P(-0.5<Z<1) = 0.5328.

Hence, there's a 53.28% chance that the demand is between 230 and 245.

c) Find the probability that demand greater than 250. (2 points)

Just like above, we'd calculate the Z values and P(Z) for 250.

Z₂₅₀ = 1.5

We've already calculated in a) that P(0<Z<1.5) = 0.4332

We need to find P(Z>1.5).

P(Z>1.5) = P(Z>0) - P(0<Z<1.5)

P(Z>1.5) = 0.5 - 0.4332

P(Z>1.5) = 0.0668

Hence, there's only a 6.68% chance that the demand will exceed 250.

d) Find the value of X for the lowest 25 percent. (4 points)

Let's assume the Z value at the lowest 25% is given as z₁

We have-

P(z₁<Z<0) = 0.25

From the standard normal tables we derive z₁ = -0.6745 (approx.)

Now lets calculate our X.

We know: Z = (X-μ)/σ

Substituting relevant values we get:

-0.6745 = (X - 235) / 10

-6.745 = X - 235

X = 228.255

Hence, the required X value is 228.255