PLEASE HELP!!!!

Im looking for some help on these questions. The answer that best answers the questions and helps me the most will receive brain

s344n2d4d5 [400]

Detailed solution is attached!!~

Diagram for ques. 3 is also attached!!

7 0

2 years ago

The number of "destination weddings" has skyrocketed in recent years. For example, many couples are opting to have their wedding

crimeas [40]

Answer:

Step-by-step explanation:

Mean = (29.7 + 29.4 + 31.7 + 29.0 + 29.1 + 30.5 + 29.1 + 29.8)/8 = 29.7875

Mean = 29.7875 × 1000 = $29787.5

Standard deviation = √(summation(x - mean)²/n

n = 8

Summation(x - mean)² = (29.7 - 29.7875)^2 + (29.4 - 29.7875)^2 + (31.7 - 29.7875)^2 + (29.0 - 29.7875)^2 + (29.1 - 29.7875)^2 + (30.5 - 29.7875)^2 + (29.1 - 29.7875)^2 + (29.8 - 29.7875)^2 = 5.88875

Standard deviation = √(5.88875/8

s = 0.88

s = 0.88 × 1000 = $880

a) We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0:µ ≤ 30000

For the alternative hypothesis,

H1:µ > 30000

This is a right tailed test.

b) The decision rule is to reject the null hypothesis if the significance level of 0.05 is greater than the probability value. If it is otherwise, we would fail to reject the null hypothesis.

c) Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 8

Degrees of freedom, df = n - 1 = 8 - 1 = 7

t = (x - µ)/(s/√n)

Where

x = sample mean = 29787.5

µ = population mean = 30000

s = samples standard deviation = 880

t = (29787.5 - 30000)/(880/√8) = - 0.68

We would determine the p value using the t test calculator. It becomes

p = 0.26

d) Since alpha, 0.1 < the p value, 0.26, then we would fail to reject the null hypothesis.

6 0

3 years ago

Michael earned $159 mowing lawns all summer. If he mowed 39 lawns, about how much did he make per lawn?

monitta

Divide 159 by 39 to get around $4.08

4 0

2 years ago

Probability math help! Please help :)

sukhopar [10]

Answer:

20%

Step-by-step explanation:

7 0

2 years ago

A survey report states that 70% of adult women visit their doctors for a physical examination at least once in two years. If 20

irakobra [83]

Answer:

a) 0.3921 = 39.21% probability that fewer than 14 of them have had a physical examination in the past two years.

b) 0.107 = 10.7% probability that at least 17 of them have had a physical examination in the past two years.

Step-by-step explanation:

For each women, there are only two possible outcomes. Either they visit their doctors for a physical examination at least once in two years, or they do not. The probability of a woman visiting their doctor at least once in this period is independent of any other women. This means that we use the binomial probability distribution to solve this question.

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.

70% of adult women visit their doctors for a physical examination at least once in two years.

This means that $p = 0.7$

20 adult women

This means that $n = 20$

(a) Fewer than 14 of them have had a physical examination in the past two years.

This is:

$P(X < 14) = 1 - P(X \geq 14)$

In which

$P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$

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

$P(X = 14) = C_{20,14}.(0.7)^{14}.(0.3)^{6} = 0.1916$

$P(X = 15) = C_{20,15}.(0.7)^{15}.(0.3)^{5} = 0.1789$

$P(X = 16) = C_{20,16}.(0.7)^{16}.(0.3)^{4} = 0.1304$

$P(X = 17) = C_{20,14}.(0.7)^{17}.(0.3)^{3} = 0.0716$

$P(X = 18) = C_{20,18}.(0.7)^{18}.(0.3)^{2} = 0.0278$

$P(X = 19) = C_{20,19}.(0.7)^{19}.(0.3)^{1} = 0.0068$

$P(X = 20) = C_{20,20}.(0.7)^{20}.(0.3)^{0} = 0.0008$

So

$P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.1916 + 0.1789 + 0.1304 + 0.0716 + 0.0278 + 0.0068 + 0.0008 = 0.6079$

$P(X < 14) = 1 - P(X \geq 14) = 1 - 0.6079 = 0.3921$

0.3921 = 39.21% probability that fewer than 14 of them have had a physical examination in the past two years.

(b) At least 17 of them have had a physical examination in the past two years

$P(X \geq 17) = P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$

From the values found in item (a).

$P(X \geq 17) = P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.0716 + 0.0278 + 0.0068 + 0.0008 = 0.107$

0.107 = 10.7% probability that at least 17 of them have had a physical examination in the past two years.

6 0

2 years ago