Evaluate h(x) = -2+ 9 when x = -2, 0, and 5

A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a variance of

iris [78.8K]

Answer:

The probability that the mean battery life would be greater than 533.2 minutes (in a sample of 75 batteries) is $\\ P(z>0.48) = P(x>533.2) = 0.3156$

Step-by-step explanation:

The main thing we have to take into account in this question is that we are about to find the probability of a sample mean. The distribution for sample means follows a normal distribution with mean $\\ \mu$ and standard deviation $\\ \frac{\sigma}{\sqrt{n}}$ . Mathematically

$\\ \overline{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

For values of the sample $\\ n \ge 30$ , no matter the distribution the data come from.

And the variable z follows a standard normal distribution, and, as we can remember, this distribution has a mean = 0 and a standard deviation = 1. Mathematically

$\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [1]

That is

$\\ z \sim N(0, 1)$

We have a variance of 3364. That is, a standard deviation of

$\\ \sigma^2 = 3364; \sigma = \sqrt{3364} = 58$

The population mean is

$\\ \mu = 530$

The sample size is $\\ n = 75$

The sample mean is $\\ \overline{x} = 533.2$

With all this information, we can solve the question

The probability that the mean battery life would be greater than 533.2 minutes

Using equation [1]

$\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ z = \frac{533.2 - 530}{\frac{58}{\sqrt{75}}}$

$\\ z = \frac{3.2}{\frac{58}{8.66025}}$

$\\ z = \frac{3.2}{6.69726}$

$\\ z = 0.47780$

With this value of z we can consult a cumulative standard normal table (or use some statistic program) to find the cumulative probability for z (and remember that this variable follows a standard normal distribution).

Most standard normal tables have values for z for only two decimals, so we can round the previous value for z as z = 0.48.

Then

$\\ P(z$

However, in the question we are asked for $\\ P(z>0.48) = P(x>533.2)$ . As well as all normal distributions, the standard normal distribution is symmetrical around the mean, and we have

$\\ P(z>0.48) = 1 - P(z$

Thus

$\\ P(z>0.48) = 1 - 0.68439$

$\\ P(z>0.48) = 0.31561$

Rounding to four decimal places, we have

$\\ P(z>0.48) = 0.3156$

So, the probability that the mean battery life would be greater than 533.2 minutes is (in a sample of 75 batteries) $\\ P(z>0.48) = P(x>533.2) = 0.3156$ .

5 0

4 years ago

The price for certain produce items at Greg's local supermarket are

Vaselesa [24]

72.28 i took the test it was easy bro

8 0

3 years ago

Help PLEASE!!! I really need it I really do please anyone

Tanya [424]

Answer:

5. alternate interior angles

7. angle addition postulate

4 0

3 years ago

James had a peach that was 98mm in diameter. One day he watered it with a magical solution, and it grew to 188,869 mm in diamete

Archy [21]

We know that 98mm is the 100% of the diameter of the plant before James watered with the magical solution; after he watered, the diameter of the plant grew to 188.869mm. Let $x$ be the grew in the percentage of the diameter.
Now we can establish a ratio and solve for $x$ :
$\frac{98}{100} = \frac{188.869}{x}$
$x= \frac{(188.869)(100)}{98}$
$x=192.7$ %
Now that we know the diameter of the plant grew 192.7%, we just need to divide that quantity by 100% to find how many times the diameter grew:
$\frac{192.7}{100} =1.927$

We can conclude that the diameter of the peach become approximately 1.9 times larger after James watered it.

4 0

4 years ago

Warren drove 18 miles to work. He drove home a different way and saw that he had driven a total of 41 miles that day. How many m

Sati [7]

Answer:

23

Step-by-step explanation:

41-18=23

3 0

3 years ago