I have a bag with 6 marbles numbered from 1 to 6. Mathew has a bag with 12 marbles numbered from 1 to 12. Matthew chooses one ma

Answer:

When x = 72, the z score is:

The average is 58

This z-score indicates that x= 72 is 1.273 standard deviations above the mean.

Step-by-step explanation:

Let’s consider the following details for this question: John's typing speed on a test is assumed to follow a normal distribution. Let X denote the number of words he can type in a minute. Hence, X ~ N(58, 11). It’s important to round the result to three decimal points if needed.

Provide your response below for words per minute on a typing test conducted on Sunday. The z score when x = 72 is

In this scenario, we acknowledge that the variable under consideration is represented by a normal distribution:

The formula for the z score is as follows:

By substituting values, we derive:

The mean is 58

This z-score indicates that x= 72 is 1.273 standard deviations higher than the mean.

The estimated probability that the code consists solely of odd digits is 0.02381.

Response:

Detailed explanation:

Greetings!

You have the variable

X: Area eligible for painting with a can of spray paint (feet²)

This variable is normally distributed with a mean of μ= 25 feet² and a standard deviation of δ= 3 feet²

As this variable has a normal distribution, it needs to be converted into the standard normal form to utilize tabulated cumulative probabilities.

a.

P(X>27)

The first step involves standardizing the X value using Z= (X-μ)/ δ ~N(0;1)

P(Z>(27-25)/3)

P(Z>0.67)

Having determined the Z value, you can find it in the table, but since the table includes probabilities for , the following conversion must be applied:

P(Z>0.67)= 1 - P(Z≤0.67)= 1 - 0.74857= 0.25143

b.

A sample of 20 cans was taken, and you need to ascertain the probability of averaging a coverage area of 540 feet².

The sample mean maintains the same distribution as its source variable, but its variance is influenced by sample size, thus it is normally distributed with parameters:

X[bar]~N(μ;δ²/n)

To cover 540 feet² with 20 cans, the average coverage must be approximately 540/20= 27 feet² per can.

c.

P(X≤27) = P(Z≤(27-25)/(3/√20))= P(Z≤2.98)= 0.999

d.

No, if the distribution is not normal and skewed, the normal distribution should not be applied for calculating probabilities. While the central limit theorem might approximate the sampling distribution to normal when the sample size is 30 or larger, that isn’t applicable here.

I trust this information is helpful!

To determine the answer, you need to divide 405 by 50 to discover the weight of a single coin. The calculation should appear like this:

= 8.1
The precise weight is 8.1 grams, but if you are looking for an estimate, the conclusion should be
Approximately 8 grams for a one-dollar coin

Answer:

7040

Step-by-step explanation:

C= 2000

B = 20/100 x 2000 = 400 + 2000 = 2400

A= 10/100 x 2400 = 240+2400= 2640

2000 + 2400 + 2640 = 7040