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Alenkinab [10]
3 years ago
6

A high percentage of people who fracture or dislocate a bone see a doctor for that condition. Suppose the percentage is 99%. Con

sider a sample in which 300 people are randomly selected who have fractured or dislocated a bone. (i) What is the probability that exactly five of them did not see a doctor? (4 marks) (ii) What is the probability that fewer than four of them did not see a doctor? (5 marks) (iii) What is the expected number of people who would not see a doctor? (5 marks)
Mathematics
1 answer:
marishachu [46]3 years ago
7 0

Answer:

(i) 0.15708

(ii) 0.432488

(iii) 3

Step-by-step explanation:

Given that, 99% of people who fracture or dislocate a bone see a doctor for that condition.

There is only two chance either the person having fracture or dislocation of bone will either see the doctor or not.

As per previous data, if one person got a fracture or dislocation of bone, the chance of seeing the doctor is 0.99. Assuming this chance is the same for every individual, so the total number of people having fractured or dislocated a bone can be considered as Bernoulli's population.

Let p be the probability of success represented by the chances of not seeing a doctor by any one individual having fractured or dislocated a bone.

So, p=1-0.99=0.01

According to Bernoulli's theorem, the probability of exactly r success among the total of n randomly selected from Bernoulli's population is

P(r)=\binom{n}{r}p^r(1-p)^{n-r}\cdots(i)

(i) The total number of persons randomly selected, n=400.

The probability that exactly 5 of them did not see a doctor

So, r=5 , p=0.01

Using equation (i),

P(r=5)=\binom{400}{5}(0.01)^5(1-0.01)^{400-5}

=\frac{400!}{(400-5)!\times 5!}(0.01)^5(0.99)^{395}

=0.15708

(ii) The probability that fewer than four of them did not see a doctor

=P(r

=P(r=0)+P(r=1)+P(r=2)+P(r=3)

=\binom{400}{0}(0.01)^0(0.99)^{400}+\binom{400}{1}(0.01)^1(0.99)^{399}+\binom{400}{2}(0.01)^2(0.99)^{398}+\binom{400}{3}(0.01)^3(0.99)^{397}

=0.017951+0.072527+0.146154+0.195856

=0.432488

(iii) The expected number of people who would not see a doctor

=np

=300\times 0.01

=3

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The Salk polio vaccine experiment in 1954 focused on the effectiveness of the vaccine in combating paralytic polio. Because it w
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Answer:

Step-by-step explanation:

Hello!

The variables of interest are:

X₁: Number of cases of polio observed in kids that received the placebo vaccine.

n₁= 201299 total children studied

x₁= 110 cases observed

X₂: Number of cases of polio observed in kids that received the experimental vaccine.

n₂= 200745 total children studied

x₂= 33 cases observed

These two variables have a binomial distribution. The parameters of interest, the ones to compare, are the population proportions: p₁ vs p₂

You have to test if the population proportions of children who contracted polio in both groups are different: p₂ ≠ p₁

a)

H₀: p₂ = p₁

H₁: p₂ ≠ p₁

α: 0.05

Z= \frac{(p'_2-p'_1)-(p_2-p_1)}{\sqrt{p'[\frac{1}{n_1} +\frac{1}{n_2} ]} }

Sample proportion placebo p'₁= x₁/n₁= 110/201299= 0.0005

Sample proportion vaccine p'₂= x₂/n₂= 33/200745= 0.0002

Pooled sample proportion p'= (x₁+x₂)/(n₁+n₂)= (110+33)/(201299+200745)= 0.0004

Z_{H_0}= \frac{(0.0002-0.0005)-0}{\sqrt{0.0004[\frac{1}{201299} +\frac{1}{200745} ]} }= -4.76

This test is two-tailed, using the critical value approach, you have to determine two critical values:

Z_{\alpha/2}= Z_{0.025}= -1.96

Z_{1-\alpha /2}= Z_{0.975}= 1.96

Then if Z_{H_0} ≤ -1.96 or if Z_{H_0} ≥ 1.96, the decision is to reject the null hypothesis.

If -1.96 < Z_{H_0} < 1.96, the decision is to not reject the null hypothesis.

⇒ Z_{H_0}= -4.76, the decision is to reject the null hypothesis.

b)

H₀: p₂ = p₁

H₁: p₂ ≠ p₁

α: 0.01

Z= \frac{(p'_2-p'_1)-(p_2-p_1)}{\sqrt{p'[\frac{1}{n_1} +\frac{1}{n_2} ]} }

The value of Z_{H_0}= -4.76 doesn't change, since we are working with the same samples.

The only thing that changes alongside with the level of significance is the rejection region:

Z_{\alpha /2}= Z_{0.005}= -2.576

Z_{1-\alpha /2}= Z_{0.995}= 2.576

Then if Z_{H_0} ≤ -2.576or if Z_{H_0} ≥ 2.576, the decision is to reject the null hypothesis.

If -2.576< Z_{H_0} < 2.576, the decision is to not reject the null hypothesis.

⇒ Z_{H_0}= -4.76, the decision is to reject the null hypothesis.

c)

Remember the level of significance (probability of committing type I error) is the probability of rejecting a true null hypothesis. This means that the smaller this value is, the fewer chances you have of discarding the true null hypothesis. But as you know, you cannot just reduce this value to zero because, the smaller α is, the bigger β (probability of committing type II error) becomes.

Rejecting the null hypothesis using different values of α means that there is a high chance that you reached a correct decision (rejecting a false null hypothesis)

I hope this helps!

8 0
3 years ago
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Alenkasestr [34]

Answer:

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Find the point that lies on the line described
Zepler [3.9K]
Y-3=-4(x-10)

Add 3 to make it go to the right side.

y=-4(x-10)+3

From here, you have the answer.
You change the sign for the 10 because it is in parentheses.

So, it is (10,3), or B.

6 0
3 years ago
The temperature in Anchorage, Alaska was 4 degrees Fahrenheit at 2:00 PM. The temperature dropped 2.5 degrees per hour until 9:0
VLD [36.1K]

Answer:

It was 26.5 degrees colder at midnight than at 2:00 pm

Step-by-step explanation:

We have that:

At 2:pm, it was 4 degrees.

The temperature dropped 2.5 degrees per hour until 9:00 pm.

From 2pm to 9pm, there are 7 hours. 2.5 per hour, it was 4 degrees at 2 pm.

So at 9 pm the temperature is:

4 - 2.5*7 = -13.5

The temperature dropped another 9 degrees by midnight.

At midnight, the temperature is -13.5 - 9 = -22.5 degrees

How much colder was it at midnight than at 2:00 pm?

2pm: 4 degrees

Midnight: -22.5 degrees

How much coldren

-22.5 - 4 = -26.5

It was 26.5 degrees colder at midnight than at 2:00 pm

5 0
3 years ago
Read 2 more answers
There are 30 students in a class. 10 students have a pet dog, 13 students have a pet cat, and 7 students have a pet fish. 4 stud
tankabanditka [31]

Answer:

9 students have pets

Step-by-step explanation:

From the above question, we are given the following information

Total number of students = 30

Let Pet Dog = D

Pet Cat = C

Pet Fish = F

Number is students that have pet dog

(D) = 10 students

Number of students that have pet cat (C) = 13 students

Number of students that have pet fish (F) = 7 students

Number of students that have Pet dog and cat ( D and C) = 4 students

Number of students that have Pet cat and fish (C and F) = 6 students

Number of student that has pet dog and pet fish (D and F) = 2 students

1 student has all three = ( D and F and C)

Number of student that have a pet Dog only

= n(D) - [n( D and C) + n( D and F) - n(D and C and F)]

= 10 -( 4+ 2 -1)

= 10 - 5

= 5

Number of student that have Pet cat only

= n(C) -[ n( D and C) + n( C and F) - n( D and C and F)]

= 13 -( 4 + 6 - 1)

= 13 - 9

= 4

Number is student that have a pet fish only

= n(F) - [n (C and F) + n( D and F) - n( D and C and F)]

= 7 - [6 + 2 - 1]

= 7 - 7

= 0

The number of students that have pets is calculated as:

(Number of students that have dogs only + Number of student that have cats only + Number of students that have fish only)

= 5 + 4 + 0

= 9

Therefore only 9 students have pets.

4 0
3 years ago
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