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emmainna [20.7K]
4 years ago
13

Students in a statistics class are conducting a survey to estimate the mean number of units students at their college are enroll

ed in. The students collect a random sample of 49 students. The mean of the sample is 12.2 units. The standard deviation is 1.6 units. What is the 95% confidence interval for the number of units students in their college are enrolled in? Assume that the distribution of individual student enrollment units at this college is approximately normal.
Mathematics
1 answer:
Katena32 [7]4 years ago
8 0

Answer:

( 11.7 , 12.7 )

Step-by-step explanation:

The provided information is:

Sample mean = \bar{X} = 12.2

Standard deviation = \sigma=1.6

Sample size = n = 49

Significance level = \alpha=0.05

Thus, the 95% confidence interval is:

\bar{X}\pm Z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\\=12.2\pm Z_{0.05/2}\frac{1.6}{\sqrt{49}}\\=12.2\pm 1.96\times0.2286\\=12.2\pm 0.4480\\=(11.7.12.7)

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<u>Number of terms = 48</u>

<u>common difference = 1.5</u>

This question involves the concept of Arithmetic Progression.

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where;

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Sum of A.P; S_{n} = 1920

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1920 = \frac{n}{2}(4 + 76)

simplifying this gives;

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  • Formula for nth term of an AP is;

t_{n} = a_{1} + (n - 1)d

where;

a_{1} is first term

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n is number of term

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the 48th term is 76

Thus;

76 = 4 + (48 - 1)d

76 - 4 = 47d

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Read more at; brainly.com/question/16935540

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