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1 answer:
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Answer:
a)
b) From the central limit theorem we know that the distribution for the sample mean is given by:
c)
Step-by-step explanation:
Let X the random variable the represent the scores for the test analyzed. We know that:
And we select a sample size of 64.
The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".
Part a
For this case the mean and standard error for the sample mean would be given by:
Part b
From the central limit theorem we know that the distribution for the sample mean is given by:
Part c
For this case we want this probability:
And we can use the z score defined as:
And using this we got:
And using a calculator, excel or the normal standard table we have that:
Answer:
0.0326 = 3.26% probability that a randomly selected thermometer reads between −2.23 and −1.69.
The sketch is drawn at the end.
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score of a measure X is given by:
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.
Mean of 0°C and a standard deviation of 1.00°C.
This means that
Find the probability that a randomly selected thermometer reads between −2.23 and −1.69
This is the p-value of Z when X = -1.69 subtracted by the p-value of Z when X = -2.23.
X = -1.69
has a p-value of 0.0455
X = -2.23
has a p-value of 0.0129
0.0455 - 0.0129 = 0.0326
0.0326 = 3.26% probability that a randomly selected thermometer reads between −2.23 and −1.69.
Sketch:
