amit works out at a gym for three quarters of an hours.he uses the treadmill for half the total workout session time and one thi
2 answers:
Answer:
Amit spends 7.5 minutes on other miscellaneous exercises.
Step-by-step explanation:
Amit works out at a gym for three quarters of an hours.
Three quarter of an hour means minutes
He uses the treadmill for half the total workout session time, means for
He uses one third time on cycle means minutes
Now, time spent on miscellaneous exercises = minutes
Therefore, total time Amit spends on other miscellaneous exercises is 7.5 minutes.
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Answer:
1) 22.66%
2) 20
Step-by-step explanation:
The scores of a test are normally distributed.
Mean of the test scores = u = 22
Standard Deviation = = 4
Part 1) Proportion of students who scored atleast 25 points
Since, the test scores are normally distributed we can use z scores to find this proportion.
We need to find proportion of students with atleast 25 scores. In other words we can write, we have to find:
P(X ≥ 25)
We can convert this value to z score and use z table to find the required proportion.
The formula to calculate the z score is:
Using the values, we get:
So,
P(X ≥ 25) is equivalent to P(z ≥ 0.75)
Using the z table we can find the probability of z score being greater than or equal to 0.75, which comes out to be 0.2266
Since,
P(X ≥ 25) = P(z ≥ 0.75), we can conclude:
The proportion of students with atleast 25 points on the test is 0.2266 or 22.66%
Part 2) 31st percentile of the test scores
31st percentile means 31%(0.31) of the students have scores less than this value.
This question can also be done using z score. We can find the z score representing the 31st percentile for a normal distribution and then convert that z score to equivalent test score.
Using the z table, the z score for 31st percentile comes out to be:
z = -0.496
Now, we have the z scores, we can use this in the formula to calculate the value of x, the equivalent points on the test scores.
Using the values, we get:
Thus, a test score of 20 represent the 31st percentile of the distribution.
Answer:
(E) 0.83
Step-by-step explanation:
We will solve it using conditional probability.
Let A be the event that a TV show is successful.
P(A) = 0.5
A' be event that the show is unsuccessful
P(A') =0.5
Let B be the event that the response was favorable
P(B) = 0.6
Let B' be the event that the response was unfavorable/
P(B') = 0.4
P(A∩B) = 0.5 and P(A∩B') = 0.3
We need to find new show will be successful if it receives a favorable response.
P(A/B) =
= 0.5/0.6
= 0.833