Alexa is the popular virtual assistant developed by Amazon. Alexa interacts with users using artificial intelligence and voice recognition. It can be used to perform daily tasks such as making to-do lists, reporting the news and weather, and interacting with other smart devices in the home. In , the Amazon Alexa app was downloaded some 2800 times per day from the Google Play store (AppBrain website). Assume that the number of downloads per day of the Amazon Alexa app is normally distributed with a mean of 2800 and standard deviation of 860. What is the probability there are 2000 or fewer downloads of Amazon Alexa in a day (to 4 decimals)? What is the probability there are between 1500 and 2500 downloads of Amazon Alexa in a day (to 4 decimals)? What is the probability there are more than 3000 downloads of Amazon Alexa in a day (to 4 decimals)? Assume that Google has designed its servers so there is probability 0.01 that the number of Amazon Alexa app downloads in a day exceeds the servers' capacity and more servers have to be brought online. How many Amazon Alexa app downloads per day are Google's servers designed to handle (to the nearest whole number)?

Assume that Google has designed its servers so there is probability 0.01 that the number of Amazon Alexa app downloads in a day exceeds the servers' capacity and more servers have to be brought online. How many Amazon Alexa app downloads per day are Google's servers designed to handle (to the nearest whole number)?

Use a chart or calculator to find the z-score given the probability.

P(Z > z) = 0.01

P(Z < z) = 0.99

z = 2.3263

Now find the corresponding number for that z-score.

z = (x − μ) / σ

2.3263 = (x − 2800) / 860

x = 4801

fichoh fichoh · Answer 2 · 2021-10-07T09:33:06+02:00

Using the binomial probability principle, the probability values for the number of downloads are :

P(X ≤ 2000) = 0.1762

P(1500 ≤ X ≤ 2500) = 0.2981

P(X > 3000) = 0.4081

Maximum number of downloads = 4800

Given the Parameters :

Mean, μ = 2800

Standard deviation, σ = 860

The Zscore formula ::

Zscore = (X - μ) / σ

1.) Probability of 2000 or fewer downloads :

P(X ≤ 2000) = P(Z ≤ (2000 - 2800)/860)) = -0.930

Using a normal distribution table :

P(Z ≤ - 0.930) = 0.1762

2.) Probability of between 1500 and 2000 downloads :

P(1500 ≤ X ≤ 2500) = P(Z ≤ (2500 - 2800)/860) - P(Z ≤ (1500 - 2800)/860)

P(1500 ≤ X ≤ 2500) = P(Z ≤ - 0.349) - P(Z ≤ - 1.511)

P(1500 ≤ X ≤ 2500) = 0.3635 - 0.0654

P(1500 ≤ X ≤ 2500) = 0.2981

3.) Probability of more than 3000 downloads :

P(X > 3000) = P(Z > (3000 - 2800) / 860)) = 0.2325

P(Z > 0.2325) = 1 - P(Z < 0.2325)

Using a normal distribution table :

P(Z > 0.2325) = 1 - 0.5919 = 0.4081

4.) Number of downloads given a probability of 0.01

P(X > 0.01) ;

Using a normal distribution table :

The Z score corresponding to P(Z > 0.01) is 2.326

Using the Zscore formula to calculate the number of downloads, X ;

Zscore = (X - μ) / σ

2.326 = (X - 2800) / 860

2.326 × 860 = X - 2800

2000.36 = X - 2800

X = 2000.36 + 2800

X = 4800.36

Therefore, maximum daily download system is designed to handle 4800.

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