Question

Question 1 The coffee shop A coffee shop knows from past records that its weekly takings (sales) are normally distributed with a mean of $10,500 and a standard deviation of $478. Answer the following questions:

1. Find the probability that in a given week the coffee shop would have takings of more than $10,700
2. Find the probability that in a given week the takings are between $9,800 and $11,000.
3. Calculate the inter-quartile range of weekly takings.
4. What are the maximum weekly takings for the worst 5% of weeks?

Question 2 Normal model

1. A cut-off score of 79 has been established for a sample of scores in which the mean is 67. If the corresponding z-score is 1.4 and the scores are normally distributed, what is the standard deviation?
2. The standard deviation of a normal distribution is 12 and 95% of the values are greater than 6. What is the value of the mean?
3. The mean of a normal distribution is 130, and only 3% of the values are greater than 155. What is the standard deviation?

Answer 1

Solution

Back-up Theory

If a random variable X ~ N(µ, σ²), i.e., X has Normal Distribution with mean µ and variance σ², then,

Z = (X - µ)/σ ~ N(0, 1), i.e., Standard Normal Distribution ….................................…………..…………..(1)

Equivalently, Z-score corresponding to X is: (X - µ)/σ ……...................................…….…………….. (1a)

and hence X corresponding to Z-score is: X = µ + Zσ ………………....................................………….. (1b)

P(X ≤ or ≥ t) = P[{(X - µ)/σ} ≤ or ≥ {(t - µ)/σ}] = P[Z ≤ or ≥ {(t - µ)/σ}] .…...........................................……(2)

Probability values for the Standard Normal Variable, Z, can be directly

read off from Standard Normal Tables ……………………………………………..…........................…… (3a)

or can be found using Excel Function: Statistical, NORMSDIST(z) which gives P(Z ≤ z) ……..........…(3b)

Percentage points of N(0, 1) can be found using Excel Function: Statistical, NORMSINV,

which gives values of t for which P(Z ≤ t) = given probability…………..............................……. ………(3c)

Now, to work out the solution,

Q1

Let X = weekly takings ($) of the coffee shop. We are given, X ~ N(10500, 478) ................................ (4)

Part (1)

Probability that in a given week the coffee shop would have takings of more than $10,700

= P(X > 10700)

= P[Z > {(10700 - 10500)/478}] [vide (2) and (4)]

= P(Z > 0.4184)

= 0.3379 [vide (3b)] Answer 1

Part (2)

Probability that in a given week the takings are between $9,800 and $11,000

P(9800 < X < 11000)

= P[{(9800 - 10500)/478} < Z < {(11000 - 10500)/478}] [vide (2) and (4)]

= P(- 1.4644 < Z < 1.0460)

= P(Z < 1.0460) - P(Z < - 1.4644)

= 0.8522 – 0.0715 [vide (3b)]

= 0.7807 Answer 2

Part (3)

Inter-quartile range {(third quartile Q3) – (first quartile Q1)}

We should have, by definition of Q1 and Q3,

P(X < Q1) = P(X > Q3) = 0.25

Or, P[Z < {(Q1 - 10500)/478}] = P[Z > {(Q3 - 10500)/478}] = 0.25 [vide (2) and (4)]

Vide (3c), {(Q1 - 10500)/478} = - 0.6745 and {(Q3 - 10500)/478} = 0.6745

Or, Q3 – Q1 = 2 x 0.6745 x 478 = 644.822. Thus,

Inter-quartile range of weekly takings = $644.82 Answer 3

Part (4)

Let the maximum weekly takings for the worst 5% of weeks be t. Then, we should have:

P(X < t) = 0.05

Or, P[Z < {(t - 10500)/478}] = 0.05 [vide (2) and (4)]

=> vide (3c), {(t - 10500)/478} = - 1.6449

Or, t = 9713.74. Thus,

maximum weekly takings for the worst 5% of weeks = $9713.74 Answer 4

Q2

Part (1)

Here we have: x = 79, mean = 67 and z-score = 1.4. So, vide (1a),

{(79 – 67)/σ} = 1.4

Or, σ = 12/1.4

= 8.57

So, standard deviation = 8.57 Answer 5

Part (2)

Here we have: σ = 12 ......................................................................................................................................... (5)

and

P(X > 6) = 0.95 =>

P[Z > {(6 - µ)/12}] = 0.05 [vide (2) and (5)

=> vide (3c), {(6 - µ)/12} = - 1.6449

Or, µ = 6 + (1.6449 x 12)

= 25.74.

Hence, mean = 25.74 Answer 6

Part (3)

Here we have: µ = 130 ......................................................................................................................................... (6)

and

P(X > 155) = 0.03 =>

P[Z > {(155 - 130)/σ}] = 0.03 [vide (2) and (6)

=> vide (3c), (25/σ) = 1.8808

Or, σ = 25/1.8808

= 13.29

So, standard deviation = 13.29 Answer 7  

DONE