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Artemon [7]
3 years ago
6

steve runs a car wash and detail shop. he determines that 50% of customers get their car washed and vacuumed, while 72% get thei

r car washed. what is the probability that a customer will get their car vacuumed given that they are getting it washed? (in decimal form rounded to nearest thousandths)​
Mathematics
1 answer:
3241004551 [841]3 years ago
8 0

Answer:

The probability is 0.694

Step-by-step explanation:

Let the probability of getting a car washed be

P(W) = 72% = 0.72

Let the probability of getting a car vacuumed be

P(V) = ?

Probability of getting car washed and vacuumed is P(W n V) = 50% = 0.5

From Bayes’ theorem

P(V|W) is the probability that a customer gets their car vacuumed given that they are getting it washed

So mathematically;

P(V|W) = P(V n W)/P(W)

= 0.5/0.72 = 0.694

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Summarize rigid body simulation.

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3 years ago
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We learned in Exercise 4.18 that about 90% of American adults had chickenpox before adulthood. We now consider a random sample o
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Answer:

a) 108 people with a standard deviation of 3.286335

b) No

c) 0.218163 or around 21.82%

d) See explanation below.

Step-by-step explanation:

This situation can be modeled with the Binomial Distribution which gives t<em>he probability of an event that occurs exactly k times out of n, and is given by </em>

<em>\large P(k;n)=\binom{n}{k}p^kq^{n-k} </em>

<em>where  </em>

<em>\large \binom{n}{k}= combination of n elements taken k at a time. </em>

<em>p = probability that the event (“success”) occurs once </em>

<em>q = 1-p </em>

In this case, the event “success” is finding an American adult who had  chickenpox before adulthood with probability p=0.9

and n=120 American adults in the sample.

The <em>standard deviation for this binomial distribution</em> is

\large s=\sqrt{npq}

where <em>n is the sample size 120 </em>

a)

We consider a random sample of 120 American adults. How many people in this sample would you expect to have had chickenpox in their childhood?

If about 90% of American adults had chickenpox before adulthood, we expect to find 90% of 120 = 0.9*120=108 people in the sample who had chickenpox in their childhood.

The  standard deviation would be

\large s=\sqrt{120*0.9*0.1=3.286335}

b)

Would you be surprised if there were 105 people who have had chickenpox in their childhood?

No, because 105 and 108 are in the interval [mean - s, mean +s]

c)

What is the probability that 105 or fewer people in this sample have had chickenpox in their childhood?

The probability that 105 or fewer people in this sample have had chickenpox in their childhood is

\large \sum_{k=0}^{105}\binom{120}{k}0.9^k0.1^{120-k}

We compute this value easily with a spreadsheet and we get

\large \sum_{k=0}^{105}\binom{120}{k}0.9^k0.1^{120-k}=0.218163\approx 21.82\%

d)

How does this probability relate to your answer to part (b)?

A binomial distribution with np>5 and nq>5 with n the sample size as is the case here, behaves pretty much like a Normal distribution with mean np and standard deviation \large \sqrt{npq}, so around 60% of the data are in the interval  [mean -s, mean +s] and 40% outside, so roughly <em>20% of the data should be in [0, mean-s] </em>

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Find the derivative of f(x) = negative 9 divided by x at x = -4. (6 points) 4 divided by 9 16 divided by 9 9 divided by 16 9 div
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