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wariber [46]
3 years ago
11

Jonathan, a college business student wants to study how many pages of notes fellow classmates took during a semester for a finan

ce class. Past data reveals that pages of notes for a finance college class has a mean of 95 , with standard deviation 25 pages. He plans to take a random sample of 30 such college students and will calculate the mean pages of notes they take to compare to the known pages of notes students take for a finance college class. For these values, the mean and standard deviation of the sampling distribution of sample means for a sample of size 30 are: μx¯=95 and σx¯=2530√=4.6. What is the probability that the sample mean for a sample of size 30 will be at least 99? You may use a calculator or the portion of the z-table given below.
Mathematics
1 answer:
Naya [18.7K]3 years ago
8 0

Answer:

I just woke up and broke my right knee

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So basically all you have to do is find the area of one of the smaller semi circles by using the formula for the area of a circle (A=πr^2). You know that the length of the larger semi circle's radius is equivalent to 6 cm because the radius of the smaller ones are 3 cm, meaning the diameter would have to be 6 cm and in this case, the length of the smaller semi circles' diameters is equal to the radius of the big semi circle. Then you would find the area of the big semi circle again by using the area of a circle formula, but after getting the answer you would half it, obviously because it's a semi circle. Subtract the are of the smaller semi circle you found earlier from the answer you just got and that's it ;) (you wouldn't have to half the area since there are two smaller semi circles and 1/2 + 1/2 = 1 but u knew that)

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4 years ago
A software developer wants to know how many new computer games people buy each year. Assume a previous study found the standard
Marrrta [24]

Answer:

The minimum sample size required to ensure that the estimate has an error of at most 0.14 at the 95% level of confidence is n=567.

Step-by-step explanation:

We have to calculate the minimum sample size n needed to have a margin of error below 0.14.

The critical value of z for a 95% confidence interval is z=1.96.

To do that, we use the margin of error formula in function of n:

MOE=\dfrac{z\cdot \sigma}{\sqrt{n}}\\\\\\n=\left(\dfrac{z\cdot \sigma}{MOE}\right)^2=\left(\dfrac{1.96\cdot 1.7}{0.14}\right)^2=(23.8)^2=566.42\approx 567

The minimum sample size to have this margin of error is n = 567.

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Step-by-step explanation:

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