Question

You may need to use the appropriate appendix table to answer this 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 2,100 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.)

downloads per day

Answer 1

Answer)

As the population is normally distributed, we can use standard normal z table to answer the problem.

Mean = 2800

S.d = 860

A)

P(x<2100)

First we need to estimate the z score,

Z = (x-mean)/s.d

Z = (2100-2800)/860

Z = -0.8140

From z table, P(z<-0.8140) = 0.2078

Required probability = 0.2078

B)

P(1400<x<2600) = P(x<2600) - p(x<1400)

P(x<2600)

Z = (2600-2800)/860 = -0.2326

From z table, P(Z<-0.2326) = 0.4081

P(x<1400)

Z = (1400-2800)/860

Z = -1.6279

From z table, P(z<-1.6279) = 0.0517

Required probability = 0.4081-0.0517

= 0.3564

C)

P(x>3100)

Z = (3100-2800)/860

Z = 0.3488

From z table, P(z<0.3488) = 0.6363

As the standard normal table is symmetrical

Therefore

P(z>0.12) = 1-0.0.6363

= 0.3636

D)

Let the number be x

First we need to estimate the corresponding z score

Z table shows P(Z<z)

That is to get 0.02 above some value

We need to look for 0.98 in the z table

P(Z<z) = 0.98

So then we will have a z score above which there would be 0.02

Z = 2.05

Z = (x-mean)/s.d

2.05 = (x-2800)/860

X = 4563