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zimovet [89]
3 years ago
10

The nursing department of a college surveyed two hundred graduates from their programs about their current

Mathematics
1 answer:
mel-nik [20]3 years ago
3 0

Answer: The required probability is 0.008.

Step-by-step explanation:

Since we have given that

Probability of bachelor's degree = 0.45

Probability of working in nursing = 0.85

Probability of both = 0.4

So, Probability of getting a graduate is currently working in nursing, given that they earned a bachelor's degree would be :

P(N|B)=\dfrac{P(N\cap B)}{P(B)}\\\\P(N|B)=\dfrac{0.4}{0.45}\\\\P(N|B)=0.008

Hence, the required probability is 0.008.

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No it is not

Step-by-step explanation:

One cannot have two outputs

Please mark me brainlest!!

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seraphim [82]

Answer:  The second one is correct I believe

Step-by-step explanation:

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Suppose the area that can be painted using a single can of spray paint is slightly variable and follows a nearly normal distribu
Svetllana [295]

Answer:

Step-by-step explanation:

Hello!

You have the variable

X: Area that can be painted with a can of spray paint (feet²)

The variable has a normal distribution with mean μ= 25 feet² and standard deviation δ= 3 feet²

since the variable has a normal distribution, you have to convert it to standard normal distribution to be able to use the tabulated accumulated probabilities.

a.

P(X>27)

First step is to standardize the value of X using Z= (X-μ)/ δ ~N(0;1)

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

P(Z>0.67)

Now that you have the corresponding Z value you can look for it in the table, but since tha table has probabilities of P(Z, you have to do the following conervertion:

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

b.

There was a sample of 20 cans taken and you need to calculate the probability of painting on average an area of 540 feet².

The sample mean has the same distribution as the variable it is ariginated from, but it's variability is affected by the sample size, so it has a normal distribution with parameners:

X[bar]~N(μ;δ²/n)

So the Z you have to use to standardize the value of the sample mean is Z=(X[bar]-μ)/(δ/√n)~N(0;1)

To paint 540 feet² using 20 cans you have to paint around 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 skewed and not normal, you cannot use the normal distribution to calculate the probabilities. You could use the central limit theorem to approximate the sampling distribution to normal if the sample size was 30 or grater but this is not the case.

I hope it helps!

4 0
3 years ago
On
Georgia [21]

Answer:

I think the first one

Step-by-step explanation: can i please get brainliest

5 0
3 years ago
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