calculate the means and the standard deviation of the following set of data 4.578g, 4.581g, 4.572g, 4.573g, 4.601g, 4.577g. stat
1 answer:
Answer:
68% Confidence interval = [4.5752, 4.5848]
95% Confidence interval = [4.5688, 4.5918]
Step-by-step explanation:
Sample mean (X) = 4.580
Sample Standard Deviation (S) = 0.01065
Sample size (n) = 6
for alpha/2 0.84 = 1.1037
for alpha/2 0.975 = 2.5706
68% Confidence interval = = [4.5752, 4.5848]
95% Confidence interval = = [4.5688, 4.5918]
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Answer:
The minimum score needed to receive a grade of A is 87.74.
Step-by-step explanation:
We are given that a professor at a local university noted that the grades of her students were normally distributed with a mean of 78 and a standard deviation of 10.
The professor has informed us that 16.6 percent of her students received grades of A.
<u><em /></u>
<u><em>Let X = grades of the students</em></u>
SO, X ~ Normal()
The z-score probability distribution for normal distribution is given by;
Z = ~ N(0,1)
where, = mean grades = 78
= standard deviation = 10
The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.
<u>Now, we are given that the professor has informed us that 16.6 percent of her students received grades of A, so the minimum score needed to receive grade A is given by;</u>
P(X x) = 0.166 {where x is the required minimum score needed}
P(
) = 0.166
P(Z
) = 0.166
<em>So, the critical value of x in the z table which represents the top 16.6% of the area is given as 0.9741, that is;</em>
<em> </em>
<em> </em> = 78 + 9.741 = <u>87.74</u>
Hence, the minimum score needed to receive a grade of A is 87.74.