10 points. See picture

The web logs of a certain website show that the average number of hits in an hour is 75 with a standard deviation equal to 8.6.

Wittaler [7]

Answer:

a) There is a 10.75% probability of observing less than 60 hits in an hour.

b) The 99th percentile of the distribution of the number of hits is 95.21 hits.

c) There is a 24% probability of observing between 80 and 90 hits an hour

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. The sum of the probabilities is decimal 1. So 1-pvalue is the probability that the value of the measure is larger than X.

In this problem, we have that

The web logs of a certain website show that the average number of hits in an hour is 75 with a standard deviation equal to 8.6, so $\mu = 75, \sigma = 8.6$ .

a) What’s the probability of observing less than 60 hits in an hour? Use the normal approximation

This is the pvalue of Z when X = 60. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 75}{8.6}$

$Z = -1.74$

$Z = -1.74$ has a pvalue of 0.1075. This means that there is a 10.75% probability of observing less than 60 hits in an hour.

b) What’s the 99th percentile of the distribution of the number of hits?

What is the value of X when Z has a pvalue of 0.99.

Z = 2.35 has a pvalue of 0.99

So

$Z = \frac{X - \mu}{\sigma}$

$2.35 = \frac{X - 75}{8.6}$

$X - 75 = 20.21$

$X = 95.21$

The 99th percentile of the distribution of the number of hits is 95.21 hits.

c) What’s the probability of observing between 80 and 90 hits an hour?

This is the pvalue of the zscore of X = 90 subtracted by the pvalue of the zscore of X = 80.

For X = 90

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{90 - 75}{8.6}$

$Z = 1.74$

$Z = 1.74$ has a pvalue of 0.95907

For X = 80

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{80 - 75}{8.6}$

$Z = 0.58$

$Z = 0.58$ has a pvalue of 0.71904

So

There is a 0.95907 - 0.71904 = 0.24003 = 24% probability of observing between 80 and 90 hits an hour

6 0

3 years ago

Find the probability of drawing a king from a standard deck of cards and then drawing a queen after the first card is replaced i

olga nikolaevna [1]

Answer:

1/169

Step-by-step explanation:

In a standard deck of cards, there are 4 kings and 4 queens, and there is a total of 52 cards. Probability is defined by the formula the number of desirable outcomes over the total outcomes. In the first draw, we have 52 total outcomes and 4 desirable (the 4 kings) that equal the fraction 4 / 52 or 1/13. We can do the same to find the probability of picking a queen next with replacement. That would equal 1/13 as well. The two events are independent because you are drawing the cards with replacement. Using the probability of independent events, the product of the probabilities at each stage is the answer: 1/13 * 1/13 = 1/169

8 0

3 years ago

A survey reveals that the sales of smartphones is increasing considerably. The average annual sales of smartphones, in million u

EleoNora [17]

Answer:

when the survey was initially conducted the average sales of smartphones was 1.7 million units

Step-by-step explanation:

The average annual sales of smartphones is given by function

$P(n)= 1.7 (b)^n$