company. Your company markets and sells two levels of network access: a resident

Question

company. Your company markets and sells two levels of network access: a residential service which includes a 50 Mbps download / 20 Mbps upload Internet connection, and a business service package which includes a 100 Mbps download / 60 Mbps upload Internet connection. Your Regional Vice President has asked for your help with some analysis of the network in connection with an annual review of your company's marketing and advertising claims. You asked your best network administrator to provide you with the mean and standard deviation of the download and upload speeds for both your resident and business class Internet services. They have confirmed that your network speeds are all approximately normally distributed and provided the following data based on a complete census of your customer data: Performance Metric Residential Service Business Class Internet Service (50 Mbps download /20 Mbps upload) (100 Mbps download/ 60 Mbps upload) Upload 61.5 Download oa Download Mean Standard Deviation (? 56.7 21.7 105.0 2.3 0.8 3.2 1.7 (8 pts) Answer the following questions about one area of town that is particularly prone to competition for Internet services where your company has 125 residential Internet customers and 87 business customers. a. What is the probability that any single residential customer would experience download speeds 1. below 57 Mbps? (Include a shaded probability plot with your answer.) What type of distribution would be used to describe the average value of download speeds of all 125 residential customers in this area? Using correct mathematical symbols, what would the mean and standard deviation of this new distribution be? (No graph needed.) What is the probability that the average download speed of all 125 residential customers in this area is less than 57 Mbps? (Include a shaded probability plot with your answer.) b. c.

Explanation / Answer

Solution

Let X = Download speeds (in Mbps) for residential internet services.

We are given X ~ N(56.7, 2.32)

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)

X bar ~ N(µ, ?2/n),…………………………………………………………….…….(3),

where X bar is average of a sample of size n from population of X.

So, P(X bar ? or ? t) = P[Z ? or ? {(?n)(t - µ)/? }] …………………………………(4)

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

Standard Normal Tables or can be found using Excel Function ……………………..(5)

If X ~ B(n, p). i.e., X has Binomial Distribution with parameters n and p, where

n = number of trials and p = probability of one success, then, probability mass function (pmf) of X is given by p(x) = P(X = x) = (nCx)(px)(1 - p)n – x, x = 0, 1, 2, ……. , n …..(6)

[The above probability can also be directly obtained using Excel Function of Binomial Distribution: BINOMDIST(Number_s:Trials:Probability_s:Cumulative), what is within brackets is (x:n:p:True)] ………………………………………………………….(6a)

Part (a)

Probability any single residential customer would experience download speeds less than 57 Mbps

= P(X < 57)

= P[Z < {(57 – 56.7)/2.3}] [vide (2)]

= P(Z < 0.1304)

= 0.5519 ANSWER 1

Part (b)

Let Xbar be the average download speed experienced by all 125 residential customers.

Then, vide (3), Xbar ~ N[56.7, (2.3/?125) ANSWER 2

i.e., the average download speed experienced by all 125 residential customers is Normally Distributed with µ = 56.7 and ? = 2.3/?125 ANSWER 3

Part (c)

Probability that the average download speed experienced by all 125 residential customers less than 57 Mbps

= P(Xbar < 57)

= P[Z < {(57 – 56.7)/0.2057}] [vide (2)]

= P(Z < 1.4584)

= 0.9276 ANSWER 4

Part (d)

Let Y = Number of customers out of the 125 residential customers who experienced download speed less than 57 mbps. Then, Y ~ B(125, p), where p = Probability any single residential customer would experience download speeds less than 57 Mbps = 0.5519 [from Part (a) above]

So, probability at least 50% of residential customers, i.e., 63 or more customers would experience download speed less than 57 mbps

= P(Y ? 63)

= 0.8382 [vide (6a)] ANSWER

DONE

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