No way to tell . . . . . we can't see the chart below.
It must be WAY down there where the sun don't shine.
So the rounded weights would be 3, 5, 2, 6, and 11, as when you round, you look at the number to the right of the one you want to round to. If it's 5 or larger, you add one to the spot you're rounding to, and make the rest of the numbers after into zeroes. If it's below 5, just change the ones after the one you want to round to into zeroes and don't add one.
Then for the estimating of the weight, you would add them together. 3+5+2+6+11. That would equal 27, so the estimate weight is 27 grams.
Answer:
(A) 0.15625
(B) 0.1875
(C) Can't be computed
Step-by-step explanation:
We are given that the amount of time it takes for a student to complete a statistics quiz is uniformly distributed between 32 and 64 minutes.
Let X = Amount of time taken by student to complete a statistics quiz
So, X ~ U(32 , 64)
The PDF of uniform distribution is given by;
f(X) =
, a < X < b where a = 32 and b = 64
The CDF of Uniform distribution is P(X <= x) =
(A) Probability that student requires more than 59 minutes to complete the quiz = P(X > 59)
P(X > 59) = 1 - P(X <= 59) = 1 -
= 1 -
=
= 0.15625
(B) Probability that student completes the quiz in a time between 37 and 43 minutes = P(37 <= X <= 43) = P(X <= 43) - P(X < 37)
P(X <= 43) =
=
= 0.34375
P(X < 37) =
=
= 0.15625
P(37 <= X <= 43) = 0.34375 - 0.15625 = 0.1875
(C) Probability that student complete the quiz in exactly 44.74 minutes
= P(X = 44.74)
The above probability can't be computed because this is a continuous distribution and it can't give point wise probability.
Answer:
0.00
Step-by-step explanation:
If the national average score on a standardized test is 1010, and the standard deviation is 200, where scores are normally distributed, to calculate the probability that a test taker scores at least 1600 on the test, we should first to calculate the z-score related to 1600. This z-score is
, then, we are seeking P(Z > 2.95), where Z is normally distributed with mean 0 and standard deviation 1. Therefore, P(Z > 2.95) = 0.00