Answers
Answer: Third answer
Step-by-step explanation:
Related Questions
An art history professor assigns letter grades on a test according to the following scheme. A: Top 5% of scores B: Scores below the top 5% and above the bottom 62% C: Scores below the top 38% and above the bottom 22% D: Scores below the top 78% and above the bottom 5% F: Bottom 5% of scores Scores on the test are normally distributed with a mean of 73.3 and a standard deviation of 9.7. Find the numerical limits for a B grade. Round your answers to the nearest whole number, if necessary.
Answers
Answer:
76 ≤ B ≤ 89
Step-by-step explanation:
Mean score (μ) = 73.3
Standard deviation (σ) = 9.7
If B Scores are below the top 5% and above the bottom 62%, then:
62% ≤ B ≤ 95%
In a normal distribution, the 62nd percentile has a corresponding z-score of z = 0.305, while the 95th percentile has a corresponding z-score of z = 1.650.
The grades X1 and X2 which are the limits for a B grade are given by:
[tex]z= \frac{X-\mu}{\sigma}\\0.305= \frac{X_1-73.3}{9,7}\\X_1=76.2\\1.650= \frac{X_2-73.3}{9,7}\\X_2=89.3[/tex]
Rounding to the nearest whole number, a B grade is given to grades between 76 and 89.
Final answer:
To calculate the numerical limits for a B grade in a normally distributed set of test scores with a mean of 73.3 and a standard deviation of 9.7, we find the z-scores corresponding to the top 5% and bottom 62%. The B grade range is approximately between 70 and 89.
Explanation:
To find the numerical limits for a B grade in a normally distributed set of test scores, we must determine the z-scores that correspond to the top 5% and the bottom 38% of scores since a B grade is assigned to scores below the top 5% and above the bottom 62% (which is equivalent to being below the top 38%). Given the mean score is 73.3 and the standard deviation is 9.7, we can use z-score formulas to convert these percentages to raw scores.
For the top 5% cutoff (z = 1.645), the score is calculated as follows:
Top 5% cutoff score = mean + (z x standard deviation)
Top 5% cutoff score = 73.3 + (1.645 x 9.7)
Top 5% cutoff score = 73.3 + 15.9565
Top 5% cutoff score ≈ 89
For the top 38% cutoff (z = -0.31), which is the bottom 62%, the score is calculated as:
Bottom 62% cutoff score = mean + (z x standard deviation)
Bottom 62% cutoff score = 73.3 + (-0.31 x 9.7)
Bottom 62% cutoff score = 73.3 + (-2.997)
Bottom 62% cutoff score ≈ 70
Therefore, a B grade on this test would be for scores roughly between 70 and 89.
Before a strike prematurely ended the 1994 major league baseball season, Tony Gwynn of the San Diego Padres had 165 hits in 419 at bats, for a .394 batting average. Baseball fans argued over whether Gwynn was effectively a .400 hitter that year. Test this hypothesis
Answers
Answer:
yes, he was effectively a .400 hitter
Step-by-step explanation:
The mean rate of success ( ÿ ) is 0.394.
The total number of hits was 419.
Under table t the value at 5% confidence interval is 1.96.
Here se the asymptotic standard deviation is given as follows
se = [tex]\sqrt{\frac{y(1 - y)}{n} }[/tex]
= [tex]\sqrt{\frac{0.394(1 - 0.394)}{419} }[/tex]
= 0.02387
The confidence interval is calculated as follows:
y ± 1.96 * se
0.394 ± 1.96 * 0.02387
the answer is -0.347 or 0.4407)
meaning that the mean lies between -0.347 to 0.4407.
There is strong evidence that 0.4 lies between the above confidence interval.
This means that Gwynn was effectively a 0.4 hitter
Tony Gwynn's batting average was .394 with 165 hits in 419 at bats. He would have needed approximately 3 more hits to officially reach a .400 average. Despite his impressive performance, it is factually incorrect to label him a .400 hitter for that season.
To test the hypothesis that Tony Gwynn was a .400 hitter in 1994 before the strike ended the season, we need to see if his actual batting average could be considered close to .400 even though it was .394.
Gwynn had 165 hits in 419 at bats, which gives a batting average of 165/419, or approximately .394 when rounded to three decimal places. To have a .400 average, he would need to have 419 * .400 hits, which equals 167.6. Therefore, he needed approximately 3 more hits (rounding up from 2.6) to achieve a .400 average with the same number of at bats.
Given that batting averages are not rounded up after the season ends, stating that Gwynn was a .400 hitter would be incorrect. However, his achievement of .394 is remarkably high, and some may argue that rounding up for discussion purposes could informally suggest he was 'effectively' a .400 hitter.
In a survey of 5100 randomly selected T.V. viewers, 40% said they watch network news programs. Find the margin of error for this survey if we want 95% confidence in our estimate of the percent of T.V. viewers who watch network news programs.
Answers
Answer:
Margin of error = 0.01344
Step-by-step explanation:
Margin of error = critical value × standard deviation.
critical value for 95% confidence interval = 1.96
Standard deviation = √[(p)(q)/n]
p = 0.4, q = 1 - p = 1 - 0.4 = 0.6, n = 5100
Standard deviation = √[(0.6×0.4)/5100] = 0.00686
Margin of error = 1.96 × 0.00686 = 0.0134