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Dahasolnce [82]
3 years ago
5

During last year's basketball season, Jackson attempted

SAT
1 answer:
OLga [1]3 years ago
6 0

This question is incomplete and not well written.

Complete question:

During last year's basketball season, Jackson attempted

150 free throws of which he made

120. So far this season, he's attempted 24 and made 16. A player's free-throw percentage is defined to be the percent of free-throws made out of the number of free-throws attempted. Assuming calculations began with the start of last year's season, which of the following best approximates Jackson's overall free-throw percentage to date?

A. 67%

B. 73%

C. 78%

D. 80%

Answer:

B. 73%

Explanation:

In the question, we are told that:

A player's free-throw percentage is defined to be the percent of free-throws made out of the number of free-throws attempted

Mathematically

Percent (% ) Free throw = (Number if free throws made ÷ Number of free throws attempted)× 100

For Last year, Jackson attempted

150 free throws of which he made

120

Percent (% ) Free throw = (Number if free throws made ÷ Number of free throws attempted) × 100

Percent(%) Free throw = (120÷150) × 100

= 80%

This season, he's attempted 24 and made 16

Percent (% ) Free throw = (Number if free throws made ÷ Number of free throws attempted) × 100

Percent(%) Free throw = (16÷24) × 100

= 66.67%

Approximately = 67%

The best approximation for Jackson's overall free-throw percentage to date is

(Percent Free Throw for last season + Percent Free Throw for this season) ÷ 2

Percent Free Throw for last season

= 80%

Percent Free Throw for this season = 66.67%

Therefore, (80% + 66.67%) ÷ 2

= 146.67% ÷ 2

= 73.335%

Approximately = 73%

Therefore, Jackson's overall free-throw percentage to date is 73%

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Serum cholesterol levels of 100 hypertensive patients were recorded. Mean cholesterol level was 190 mg/d1. The data was symmetri
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The standard error is 1.5.

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<h3>What is the standard error of the distribution?</h3>

The standard error of the distribution is calculated as follows:

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where N is the number of participants = 100

From the empirical rule of a normal or symmetrical distribution;

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  • 95% percent within two standard deviations, and
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Therefore,  175 mg/dl and 205 mg/dl lie one standard deviation from the mean.

One standard deviation = 175 - 190 or 205 - 190 = ±15

Standard deviation = 15

a. Solving for the standard error using the equation above;

Standard error = 15/√100

Standard error = 1.5

b. The range of 95% confidence interval is two standard deviations away from the mean.

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Range of values two standard deviations from the mean = 160 mg/dL and 220 mg/dL

Therefore, the range of 95% confidence interval is 160 mg/dL and 220 mg/dL

In conclusion, the standard error is calculated from the standard deviation and the sample size.

Learn more about standard error at: brainly.com/question/14467769

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