HELPPPP PLEASE IK ITS LONG BUT PLEASE HELP 9-12
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Using the normal distribution, we have that:
- For a single value, P(X < 79.1) = 0.5517.
- For the sample of n = 155, P(X < 79.1) = 0.9463.
The z-score of a measure X of a normally distributed variable with mean and standard deviation
is given by:
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
- By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation
.
The mean and the standard deviation are given, respectively, by:
.
The probability is the <u>p-value of Z when X = 79.1</u>, hence:
Z = (79.1 - 76.2)/22.4
Z = 0.13
Z = 0.13 has a p-value of 0.5517.
Hence: P(X < 79.1) = 0.5517.
For the sample of 155, applying the Central Limit Theorem, the standard error is:
s = 22.4/sqrt(155) = 1.8
Hence:
Z = (79.1 - 76.2)/1.8
Z = 1.61
Z = 1.61 has a p-value of 0.9463.
P(X < 79.1) = 0.9463.
More can be learned about the normal distribution at brainly.com/question/15181104
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Answer:
a: 105.2 < µ < 112.8
b: 104.872 < µ < 113.128
c: 105.841 < µ < 112.159
d: No, because n < 30
Step-by-step explanation:
For a - c, see attached photos for work. There are 2 formulas to use. The steps for constructing any confidence interval are the same, you will just use different numbers in the formula depending on what data is given to you.
d: With large sample sizes, the data often resembles normally distributed data, so we can still construct confidence intervals from the data.
