Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

What is the z-score of x = -2, if it is 2.78 standard deviations to the left of

ID: 3324746 • Letter: W

Question

What is the z-score of x = -2, if it is 2.78 standard deviations to the left of the mean?

Less than 0

1.6 – 2.5

0 – 0.5

More than 2.5

0.6 – 1.5

11) Suppose a normal distribution has a mean of six and a standard deviation of 1.5. What is the z-score if x = 5.5? Save your answer, as you’ll need it for the next two questions.

1.6 – 2.5

0 – 0.5

0.6 – 1.5

Less than 0

More than 2.5

12) For the z score you just calculated, what’s the corresponding p value? This p value measures the percent of the area under the curve to the left of z. Hint: draw a diagram.

0% – 20%

20% – 40%

40% - 60%

60% - 80%

80% - 100%

13) And based on the z score you calculated for question 11, what’s the percent of the area to the right of z.

0% – 20%

20% – 40%

40% - 60%

60% - 80%

80% - 100%

14) In a normal distribution, x = 5 while the z score = –1.25. This tells you that x = 5 is ____ standard deviations to the ____ (right or left) of the mean.

5, right

1.25, right

5, left

1.25, left

15) If the area to the right of x in a normal distribution is .243, what is the area to the left of x?

Less than 0.45

0.56 – 0.65

More than 0.75

0.45 – 0.55

0.66 – 0.75

16) Christmas trees can be recycled by feeding them to goats. Assume that the number of trees a goat can eat is normally distributed. The mean goat can eat 20 trees an hour, with a standard deviation of 2 trees. Sixty-eight percent of the time, how many trees can a goat consume?

19-21

Exactly 20

10-30

15-25

18-22

17) Christmas trees can be recycled by feeding them to goats. Assume that the number of trees a goat can eat is normally distributed. The mean goat can eat 20 trees an hour, with a standard deviation of 2 trees. What percent of the time will a goat consume between 16 and 24 Xmas trees?

50%

0%

99.7%

95%

68%

18) The average professor nods off at his desk 384 times in any given year, with a standard deviation of 50 nods. Assume that nods are normally distributed (I assume it because I made all this up). What’s the probability that a professor nods off less than 250 times?

0.16 – 0.25

More than 0.35

0.05 – 0.15

0.26 – 0.35

Less than 0.05

19) The average professor nods off at his desk 384 times in any given year, with a standard deviation of 50 nods. Assume that nods are normally distributed (I assume it because I made all this up). What’s the probability that a professor nods off less than 420 times?

More than 0.85

0.66 – 0.75

0.76 – 0.85

0.55 – 0.65

Less than 0.55

20) The average professor nods off at his desk 384 times in any given year, with a standard deviation of 50 nods. Assume that nods are normally distributed (I assume it because I made all this up). What’s the probability that a professor nods off more than 420 times?

Less than 0.05

0.16 – 0.25

More than 0.35

0.05 – 0.15

0.26 – 0.35

21) The average professor nods off at his desk 384 times in any given year, with a standard deviation of 50 nods. Assume that nods are normally distributed (I assume it because I made all this up). What’s the probability that a professor nods off more than 1,000,000 times?

Less than 0.05

0.05-0.15

More than 0.35

0.26 – 0.35

0.16 – 0.25

Explanation / Answer

Q10) Option A is Correct. Less than zero, because it lies to the left of mean

Q11) Option D is Correct. Less than zero

z = (5.5-6)/1.5

z = -0.33

Q12) p value = P(z<-0.33)

P value = 0.3694

So, 20%-40%

Option B is Correct

Q13) Area to right of z = 1-0.3694 = 0.6306

60%-80%

Option D is Correct

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote