Question

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 2018, the Amazon Alexa app was downloaded some 2,800 times per day from the Google Play store.† Assume that the number of downloads per day of the Amazon Alexa app is normally distributed with a mean of 2,800 and standard deviation of 860.

(a)

What is the probability there are 1,900 or fewer downloads of Amazon Alexa in a day? (Round your answer to four decimal places.)

(b)

What is the probability there are between 1,400 and 2,600 downloads of Amazon Alexa in a day? (Round your answer to four decimal places.)

(c)

What is the probability there are more than 3,100 downloads of Amazon Alexa in a day? (Round your answer to four decimal places.)

(d)

Suppose that Google has designed its servers so there is probability 0.02 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? (Round your answer to the nearest integer.)

Answer 1

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.