Dr. Patton is a professor of English. Recently she counted the number of misspelled words in a group of student essays. She noted the distribution of misspelled words per essay followed the normal distribution with a population standard deviation of 2.44 words per essay. For her 10 a. m. section of 40 students, the mean number of misspelled words was 6.05. Construct a 95% confidence interval for the mean num- ber of misspelled words in the population of student essays.

The confidence interval is (5.2938, 6.8062)

Step-by-step explanation:

Given that:

Standard deviation (σ) = 2.44 words per essay,

Mean (μ) = 6.05,

Confidence interval (c) = 95% = 0.95.

number of students (n) = 40 students

α = 1 - c = 1 - 0.95 = 0.05

From the probability table,

The margin of error (e) is given by the equation:

Substituting values:

The confidence interval  is given by (μ - e, μ + e).

Therefore: The confidence interval = (μ - e, μ + e)  = (6.05 - 0.7562, 6.05 + 0.7562) = (5.2938, 6.8062)

The confidence interval is (5.2938, 6.8062)

f-1(x)= cube root of 19x squared/x

step-by-step explanation:

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step-by-step explanation:

hopefully im right