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Romashka-Z-Leto [24]
3 years ago
12

An educational psychologist wishes to know the mean number of words a third grader can read per minute. She wants to make an est

imate at the 95% level of confidence. For a sample of 695 third graders, the mean words per minute read was 38.3. Assume a population standard deviation of 3.6. Construct the confidence interval for the mean number of words a third grader can read per minute. Round your answers to one decimal place.
Mathematics
1 answer:
amid [387]3 years ago
8 0

Answer:

(38.1,88.6)

Step-by-step explanation:

We are given the following in the question:

Sample mean, \bar{x} = 38.3

Sample size, n = 695

Alpha, α = 0.05

Population standard deviation, σ = 3.6

95% Confidence interval:

\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}

Putting the values, we get,

z_{critical}\text{ at}~\alpha_{0.05} = 1.96

38.3 \pm 1.96(\frac{3.6}{\sqrt{695}} ) = 38.3 \pm 0.267 = (38.033,38.567) \approx (38.1,88.6)

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Answer:

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Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

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The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

In this question:

Mean \mu, standard deviation \sigma

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3 0
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Hope this helps!
3 0
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