An English professor assigns letter grades on a test according to the following scheme. A: Top 7% of scores B: Scores below the
top 7% and above the bottom 59% C: Scores below the top 41% and above the bottom 18% D: Scores below the top 82% and above the bottom 8% F: Bottom 8% of scores Scores on the test are normally distributed with a mean of 75.7 and a standard deviation of 8.8. Find the numerical limits for a C grade. Round your answers to the nearest whole number, if necessary.
1 answer:
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Answer:
95% confidence interval for p, the true proportion of all women who will find success with this new treatment is [0.238 , 0.762].
Step-by-step explanation:
We are given that a pharmaceutical company proposes a new drug treatment for alleviating symptoms of PMS (premenstrual syndrome).
In the first stages of a clinical trial, it was successful for 7 out of the 14 women.
Firstly, the pivotal quantity for 95% confidence interval for the true proportion is given by;
P.Q. = ~ N(0,1)
where, = sample proportion of women who find success with this new treatment =
= 0.50
n = sample of women = 14
<em>Here for constructing 95% confidence interval we have used One-sample z proportion statistics.</em>
So, 95% confidence interval for the true proportion, p is ;
P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%
level of significance are -1.96 & 1.96}
P(-1.96 < < 1.96) = 0.95
P( <
<
) = 0.95
P( < p <
) = 0.95
<u />
<u>95% confidence interval for p</u> = [ ,
]
= [ ,
]
= [0.238 , 0.762]
Therefore, 95% confidence interval for p, the true proportion of all women who will find success with this new treatment is [0.238 , 0.762].