A simple random sample of 60 is drawn from a normally distributed population, and the mean is found to be 28, with a standard de
1 answer:
The value of 27 is within 95% confidence interval because it is greater than 26.7 and less than 29.3.
<h3>What is a confidence interval?</h3>In statistics, a confidence interval is a range to estimate for unknown terms.
The most common level is the 95% confidence interval.
It can be calculated by
CI = Mean ± margin of error.
α = 1 - 0.95 = 0.05
α/2 = 0.025
z(0.025) = 1.96
We need to find the
= 0.645
So,
= 1.96 × 0.645
CI =1.26
C.I= 28±1.2652
Upper limit = 28+1.2652
=29.2625
Lower limit =28-1.2652
=26.7348
Hence, The value of 27 is within 95% confidence interval because it is greater than 26.7 and less than 29.3.
Learn more about confidence intervals;
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Answer:
a) This means that there is a 33% probability that one car chosen at random will have less than 49.5 tons of coal.
b) There is a 0.04% .probability that 35 cars chosen at random will have a mean load weight of less than 49.5 tons of coal
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
The actual weights of coal loaded into each car are normally distributed, with mean = 50 tons and standard deviation = 0.9 ton. This means that
(a) What is the probability that one car chosen at random will have less than 49.5 tons of coal? (Round your answer to four decimal places.)
This is the pvalue of Z when
has a pvalue of 0.33.
This means that there is a 33% probability that one car chosen at random will have less than 49.5 tons of coal.
(b) What is the probability that 35 cars chosen at random will have a mean load weight of less than 49.5 tons of coal? (Round your answer to four decimal places.)
Now we have to use the standard deviation of the sample, since we are working with the sample mean. That is
Now, we find pvalue of Z when
has a pvalue of 0.0004.
This means that there is a 0.04% .probability that 35 cars chosen at random will have a mean load weight of less than 49.5 tons of coal