Ty had 10 eggs. He made 4 servings of scrambled eggs. He used n eggs for each serving.

One US dollar is currently worth 0.79 euro (€). If you exchange $522 for euros, how many euros will you have?

Alecsey [184]

it should be 412.38 but thats what i got on my math

5 0

2 years ago

In a production process, the diameter measures of manufactured o-ring gaskets are known to be normally distributed with a mean d

Art [367]

Answer:

DIRECT WAY EXCEL

"=NORM.DIST(75,80,3,TRUE)"

And we got: $P(X\leq 75)= 0.04779$

OTHER WAY

$P(X\leq 75)=P(\frac{X-\mu}{\sigma}\leq \frac{75-\mu}{\sigma})=P(Z\leq \frac{75-80}{3})=P(Z$

And we can find this probability using the normal standard table or excel:

$P(Z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the diameter of a population, and for this case we know the distribution for X is given by:

$X \sim N(80,3)$

Where $\mu=80$ and $\sigma=3$

And we know that if the diameter is 75 or less the ring would be considered defective , so then in order to find the proportion of defective we need to find the following probability:

$P(X\leq 75)$

One way to do this in excel is with the following formula:

"=NORM.DIST(75,80,3,TRUE)"

And we got: $P(X\leq 75)= 0.04779$

And the other way is use the z score formula given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X\leq 75)=P(\frac{X-\mu}{\sigma}\leq \frac{75-\mu}{\sigma})=P(Z\leq \frac{75-80}{3})=P(Z$

And we can find this probability using the normal standard table or excel:

$P(Z$

3 0

2 years ago

A simple random sample is drawn from a normally distributed population, and when making a statistical inference about the popula

DiKsa [7]

To find the confidence interval, add and subtract the margin of error from the mean. With mean 18.7 and margin of error 5.9, you have 95% confidence the answer is between 12.8 and 24.6.

3 0

2 years ago

Read 2 more answers

3.52 A coin is tossed twice. Let Z denote the number of heads on the first toss and W the total number of heads on the 2 tosses.

Rufina [12.5K]

Answer:

a) The joint probability distribution

P(0,0) = 0.36, P(1,0) = 0.24, P(2,0) = 0, P(0,1) = 0, P(1,1) = 0.24, P(2,1)= 0.16

b) P( W = 0 ) = 0.36, P(W = 1 ) = 0.48, P(W = 2 ) = 0.16

c) P ( z = 0 ) = 0.6

P ( z = 1 ) = 0.4

Step-by-step explanation:

Number of head on first toss = Z

Total Number of heads on 2 tosses = W

% of head occurring = 40%

% of tail occurring = 60%

P ( head ) = 2/5 , P( tail ) = 3/5

a) Determine the joint probability distribution of W and Z

P( W =0 |Z = 0 ) = 0.6 P( W = 0 | Z = 1 ) = 0

P( W = 1 | Z = 0 ) = 0.4 P( W = 1 | Z = 1 ) = 0.6

P( W = 1 | Z = 0 ) = 0 P( W = 2 | Z = 1 ) = 0.4

The joint probability distribution

P(0,0) = 0.36, P(1,0) = 0.24, P(2,0) = 0, P(0,1) = 0, P(1,1) = 0.24, P(2,1)= 0.16

B) Marginal distribution of W

P( W = 0 ) = 0.36, P(W = 1 ) = 0.48, P(W = 2 ) = 0.16

C) Marginal distribution of Z ( pmf of Z )

P ( z = 0 ) = 0.6

P ( z = 1 ) = 0.4

8 0

2 years ago

Can you please help me fill out the blanks? I am in need of assistance. I will mark the BRAINLIEST to the first person who answe

GREYUIT [131]

I solve this question by hand and have attached images for you because it is very difficult or impossible to do it here . I do it as simple as I can and make it very comprehensive for you. Hopefully, you will understand.

Note: the order for images remain same as I have attached.

7 0

2 years ago