Continuous Probability Distributions Show all workings. Question 1 The coffee sh

Question

Continuous Probability Distributions Show all workings. Question 1 The coffee shop A coffee shop knows from past records that its weekly takings (sales) are normally distributed with a mean of $12,500 and a standard deviation of $3200. Answer the following questions: Find the probability that in a given week the coffee shop would have takings of more than $15,000 (a) (b) Find the probability that in a given week the takings are between $10,000 and $15,000. (c) (d) (e) What is the chance that the weekly takings are exactly $12,500? Calculate the inter-quartile range of weekly takings. (Hint: Locate Q1 and Q3 first.) what are the maximum weekly takings for the worst 5% of weeks? Question 2 Normal model A cut-off score of 77 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? The standard deviation of a normal distribution is 11 and 90% of the values are greater than 5. What is the value of the mean? The mean of a normal distribution is 130, and only 7% of the values are greater than 155. What is the standard deviation? (a) (c)

Explanation / Answer

Let X = weekly takings (sales in $) of the coffee shop.

We are given X ~ N(12500, 32002).

Back-up Theory

If a random variable X ~ N(µ, 2), i.e., X has Normal Distribution with mean µ and variance 2, then,

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

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

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

Standard Normal Tables or can be found using Excel Function……………………..(3)

Inter-quartile Range, IR = Q3 – Q1 …………………………………………………(4)

First Quartile Q1 of a random variable X is defined such that P(X Q1) = 0.25 …..(5) and the Third Quartile Q3 is defined such that P(X Q3) = 0.75 ……………………….(6)

Part (a)

Probability that in a given week the coffee shop would have takings of more than $15000

= P(X > 15000)

= P[Z > {(15000 - 12500)/3200}] [vide (2) of Back-up Theory]

= P(Z > 0.78125)

= 0.2173 [using Excel Function on Standard Normal Distribution] ANSWER

Part (b)

Probability that in a given week the takings of the coffee shop would be between $10000 and $15000

= P(10000 < X < 15000)

= P[{(10000 - 12500)/3200}Z < {(15000 - 12500)/3200}] [vide (2) of Back-up Theory]

= P(- 0.78125 < Z < 0.78125)

= P(Z < 0.78125) - P(Z < - 0.78125)

= 0.7827 – 0.2173[using Excel Function on Standard Normal Distribution]

= 0.5654 ANSWER

Part (c)

Chance that weekly takings are exactly $ 1200 = 0 because probability of a continuous variable taking an exact value is zero. ANSWER

Part (d)

We should have P(X Q1) = 0.25 [vide (5) of Back-up Theory]

=> P[Z {(Q1 - 12500)/3200}] = 0.25[vide (2) of Back-up Theory]

=> {(Q1 - 12500)/3200} = - 0.67449 [using Excel Function on Standard Normal Distribution]

=> Q1 = 10341.632

Similarly,

P(X Q3) = 0.75 [vide (5) of Back-up Theory]

=> P[Z {(Q3 - 12500)/3200}] = 0.75[vide (2) of Back-up Theory]

=> {(Q3 - 12500)/3200} = [using Excel Function on Standard Normal Distribution]

=> Q3 = 14658.368

Thus, inter-quartile range, IR = 14658.368 - 10341.632 = 4316.736 ANSWER

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