why did the united states push for control of the oregon country? increased american settlement in the pacific northwest increas
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Answer:
0.4514 = 45.14% probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled
Step-by-step explanation:
To solve this question, we have to understand the normal probability distribution and the central limit theorem.
Normal probability distribution:
Problems of normally distributed samples are 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.
Central limit theorem:
The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
In this problem, we have that:
What is the probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled?
This is the pvalue of Z when X = 159+1 = 160 subtracted by the pvalue of Z when X = 159-1 = 158. So
X = 160
By the Central Limit Theorem
has a pvalue of 0.7257
X = 150
has a pvalue of 0.2743
0.7257 - 0.2743 = 0.4514
0.4514 = 45.14% probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled
Answer:
Step-by-step explanation:
From the diagram which I have drawn and attached below:
Next, in the triangle, the sum of the three interior angles:
The value of angle x is 35 degrees.
Answer:
Step-by-step explanation:
From the given information; Let's assume that R should represent the set of all possible outcomes generated from a bit string of length 10 .
So; as each place is fitted with either 0 or 1
Similarly; the event E signifies the randomly generated bit string of length 10 does not contain a 0
Now;
if a 0 bit and a 1 bit are equally likely
The probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if a 0 bit and a 1 bit are equally likely is;
so ; if bits string should not contain a 0 and all other places should be occupied by 1; Then:
;