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3 years ago

Please help me! It isn’t hard im just not sure how to do it .

Mathematics

1 answer:

pishuonlain [190]3 years ago

4 0

Answer:

a. Semester 1: 86.8, Semester 2: 87

b. Semester 1: 5.7, Semester 2: 6.3

c. Semester 1 because the standard deviation of the first semester was lower than of the second. This shows that his scores were closer to the mean and therefore he was more consistent.

Step-by-step explanation: Find the mean by taking the sum of the values for each semester and divide that sum by the number of values to find the average. Do this for both semesters. Now, find the absolute value of the difference between the mean and each value of the corresponding data set. Square each if those differences and sum them up. Now divide that sum by the number of data values and take the square root. This number is your standard deviation. You will use the standard deviations to make the comparison for part c. I hope this makes sense and helps!!!!

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3 years ago

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Step-by-step explanation:

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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3 years ago

In a study, 36% of adults questioned reported that their health was excellent. A researcher wishes to study the health of peop

steposvetlana [31]

Answer:

0.3907

Step-by-step explanation:

We are given that 36% of adults questioned reported that their health was excellent.

Probability of good health = 0.36

Among 11 adults randomly selected from this area, only 3 reported that their health was excellent.

Now we are supposed to find the probability that when 11 adults are randomly selected, 3 or fewer are in excellent health.

i.e. $P(x\leq 3)=P(x=1)+{P(x=2)+P(x=3)$

Formula : $P(x=r)=^nC_r p^r q ^ {n-r}$

p is the probability of success i.e. p = 0.36

q = probability of failure = 1- 0.36 = 0.64

n = 11

So, $P(x\leq 3)=P(x=1)+{P(x=2)+P(x=3)$

$P(x\leq 3)=^{11}C_1 (0.36)^1 (0.64)^{11-1}+^{11}C_2 (0.36)^2 (0.64)^{11-2}+^{11}C_3 (0.36)^3 (0.64)^{11-3}$

$P(x\leq 3)=\frac{11!}{1!(11-1)!} (0.36)^1 (0.64)^{11-1}+\frac{11!}{2!(11-2)!} (0.36)^2 (0.64)^{11-2}+\frac{11!}{3!(11-3)!} (0.36)^3 (0.64)^{11-3}$

$P(x\leq 3)=0.390748$

Hence the probability that when 11 adults are randomly selected, 3 or fewer are in excellent health is 0.3907

5 0

3 years ago