Andrew solved the following inequality, and his work is shown below:

g The downtime per day for a computing facility has mean 4 hours and standard deviation 0.9 hour. What assumptions must be true

Sedaia [141]

Answer:

To obtain a valid approximation for probabilities about the average daily downtime, either the underlying distribution(of the downtime per day for a computing facility) must be normal, or the sample size must be of 30 or more.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

In this question:

To obtain a valid approximation for probabilities about the average daily downtime, either the underlying distribution(of the downtime per day for a computing facility) must be normal, or the sample size must be of 30 or more.

7 0

3 years ago

You are running a consention stand at a concersion stand at a basketball game. you are selling hot dogs and sodas. Each soda is

Tanya [424]

Answer:

38

Step-by-step explanation:

8 0

3 years ago

During the holidays, Duncan works as a gift wrapper. He works 35 hours a week, spread equally over 5 workdays. If Duncan earns $

Angelina_Jolie [31]

Yes my mom and mom I got a baby text you baby text me when your baby girl love mom

3 0

3 years ago

Read 2 more answers

a. Among a shipment of 5,000 tires, 1,000 are slightly blemished. If one purchases 10 of these tires, what is the probability th

hodyreva [135]

Answer:

<h2>See the explanation.</h2>

Step-by-step explanation:

3 or less than 3 tires among the 10 tires should be blemished.

There are some possibilities.

1000 tires are blemished, (5000 - 1000) = 4000 tires are not blemished.

From the 5000 tires, 10 tires can be chosen in $^{5000}C_{10}$ ways.

Possibility 1: <u>3 tires are blemished.</u>

Total 10 tires can be chosen with 3 blemished tires in $^{1000}C_3 \times {4000}C_7$ ways.

Possibility 2: <u>2 tires are blemished.</u>

Total 10 tires can be chosen with 2 blemished tires in $^{1000}C_2 \times ^{4000}C_8$ ways.

Possibility 3: <u>1 tires are blemished.</u>

Total 10 tires can be chosen with 1 blemished tires in $^{1000}C_1 \times^{4000}C_9$ ways.

Possibility 4: <u>No tires are blemished.</u>

Total 10 tires can be chosen with no blemished tires in $^{1000}C_0 \times^{4000}C_{10}$ ways.

Hence, the required probability is $\frac{^{1000}C_1 \times^{4000}C_9 + ^{1000}C_0 \times^{4000}C_{10} + ^{1000}C_2 \times^{4000}C_8 + ^{1000}C_3 \times^{4000}C_7}{^{5000}C_{10} }$ .

6 0

3 years ago

Please help and tysm!! :)) <3

kvv77 [185]

Answer:

39

Step-by-step explanation:

above 50 it's nearest 100.

Please mark me as a brainliest

3 0

2 years ago

Read 2 more answers