Can someone help me with this please

Police estimate that 25% of drivers drive without their seat belts. If they stop 6 drivers at random, find the probability tha

Furkat [3]

Answer:

17.80% probability that all of them are wearing their seat belts.

Step-by-step explanation:

For each driver stopped, there are only two possible outcomes. Either they are wearing their seatbelts, or they are not. The drivers are chosen at random, which mean that the probability of a driver wearing their seatbelts is independent from other drivers. So we use the normal probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Police estimate that 25% of drivers drive without their seat belts.

This means that 75% wear their seatbelts, so $p = 0.75$

If they stop 6 drivers at random, find the probability that all of them are wearing their seat belts.

This is $P(X = 6)$ .

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 6) = C_{6,6}.(0.75)^{6}.(0.25)^{0} = 0.1780$

17.80% probability that all of them are wearing their seat belts.

3 0

3 years ago

Read 2 more answers

Choose the correct term that makes the sentence true. The (greatest) (or) (least) of the common factors of two or more numbers i

stiv31 [10]

the greatest is 10 and the least is one but I don't know for sure but I don't know if that what you looking for

5 0

4 years ago

In recent years more people have been working past the age of 65. In 2005, 27% of people aged 65–69 worked. A recent report from

Vinil7 [7]

Answer:

a) point estimate is 30%

b) null and alternative hypothesis would be

$H_{0}$ : p=27%

$H_{a}$ : p>27%

c) We reject the null hypothesis, percentage working people aged 65-69 had increased

Step-by-step explanation:

a. 

Point estimate would be the proportion of the working people aged 65–69 to the sample size and equals $\frac{180}{600}=0.3$ ie 30%

b.

Let p be the proportion of people aged 65–69 who is working. OECD claims that percentage working had increased. Then null and alternative hypothesis would be

$H_{0}$ : p=27%

$H_{a}$ : p>27%

c.

z-score of the sample proportion assuming null hypothesis is:

$\frac{p(s)-p}{\sqrt{\frac{p*(1-p)}{N} } }$ where

p(s) is the sample proportion of working people aged 65–69 (0.3)
p is the proportion assumed under null hypothesis. (0.27)
N is the sample size (600)

then z= $\frac{0.3-0.27}{\sqrt{\frac{0.27*0.73}{600} } }$ = 1.655

Since one tailed p value of 1.655 = 0.048 < 0.05, sample proportion is significantly different than the proportion assumed in null hypothesis. Therefore we reject the null hypothesis.

3 0

3 years ago

Read 2 more answers

NEED HELP ASAP PLEASE

Tju [1.3M]

Answer:

9/35

Step-by-step explanation:

no of even number =3(2,4,6)

probablity of getting even in both dice =(3/6)×(3/6)

=9/36

4 0