Question

Please Help! 50 points!!! (Dont spam my question ill report you)

The probability of success in each of the 58 identical engine tests is p=0.92. What is the mean of this binomial distribution?


4.3

53.4

2.1

63.0


The minimum usual value of a binomial distribution is 89.3. The maximum usual value is 102.8. Which of the following values would be considered usual for this distribution?


103

89

91

125


A local poll found that only 17% of the 546 people surveyed feel the mayor's new city tax plan is effective. Find the mean of this binomial distribution.


77.0

453.2

8.8

92.8


36% of the 1035 people who watched the new sequel to a popular action movie thought it was better than the orginal. Find the mean.


372.6

238.5

662.4

15.4


A bonus activity in a popular video game allows the gamer to choose from several boxes to acquire a rare and useful item. The probability of success is 0.2. With 35 attempts, find the standard deviation.


2.4

7.0

5.6

0.2


The probability of a child being born with light-colored eyes to two particular parents is 0.25. The parents intend to have 5 children. Find the variance of this binomial distribution.


0.9

1.0

1.3

3.8


The probability of students choosing the left lunch line at school is p=0.431. Find the variance of the binomial distribution of 982 students choosing the left lunch line today.


423.2

558.8

15.5

240.8


A couple of college students, frustrated with the current class registration process, decide to survey 500 random students on campus. They find that 84% are also dissatisfied. What is the standard deviation of this binomial distribution?


420.0

8.2

67.2

0.1

Answer 1

Answer:

Explanation:

1)The probability of success in each of the 58 identical engine tests is p=0.92

n = 58

mean, u = np = 58×0.92 = 53.36

2) The only value that would be considered usual for this distribution is 91. This is because it is the only value between the minimum and maximum value

3) n = 546

p = 17/100 = 0.17

Mean = np = 546×0.17= 92.82

4) n = 1035

p = 36/100 = 0.36

np = 1035 × 0.36 = 372.6

5) The probability of success is 0.2.

p = 0.2

q= 1-p = 1-0.2 = 0.8

n = 35

standard deviation =

√npq = √35×0.2×0.8 = 2.34

6) p = 0.25

q = 1-0.25 = 0.75

n = 5

Variance = npq = 5×0.25×0.75 = 0.9

7) n = 982

p = 0.431

q = 1 - p = 1 - 0.431 = 0.569

Variance = npq = 982×0.431×0.569= 240.8

8) n = 500

p = 84/100 = 0.84

q = 1-0.84 = 0.16

Standard deviation = √npq

Standard deviation = √500×0.84×0.16 = 8.2