Can someone help me with this please???
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Answer:
The margin of error for the 95% confidence interval for the mean score of all such subjects is of 8.45.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 27 - 1 = 26
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 26 degrees of freedom(y-axis) and a confidence level of . So we have T = 2.0518
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample.
In this question:
. So
The margin of error for the 95% confidence interval for the mean score of all such subjects is of 8.45.
Using the normal distribution, it is found that the probability is 0.16.
<h3>Normal Probability Distribution</h3>In a normal distribution with mean and standard deviation
, the z-score of a measure X is given by:
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
In this problem, the mean and the standard deviation are given by, respectively, .
The proportion of students between 45 and 67 inches is the p-value of Z when <u>X = 67 subtracted by the p-value of Z when X = 45</u>, hence:
X = 67:
Z = -1
Z = -1 has a p-value of 0.16.
X = 45:
Z = -8.3
Z = -8.3 has a p-value of 0.
0.16 - 0 = 0.16
The probability is 0.16.
More can be learned about the normal distribution at brainly.com/question/24663213