Please can you answer (2.5 × 10−2) ÷ (3.8 × 103)
2 answers:
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Answer:
The answer is 105/4
Step-by-step explanation:
Answer:
b) 6.68%
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.
The mean score on the scale is 50. The distribution has a standard deviation of 10.
This means that
Matthew scores a 65. What percentage of people could be expected to score the same as Matthew or higher on this scale?
The proportion is 1 subtracted by the p-value of Z when X = 65. So
has a p-value of 0.9332.
1 - 0.9332 = 0.0668
0.0668*100% = 6.68%
So the correct answer is given by option b.
The 90% , 99% confidence interval for the population mean is 32.145 < < 35.855 and 31.093 <
< 36.907
Probability is the study of likeliness of an event to happen.
It is given that
Total Population = 50
Mean = 35
The confidence interval is given by
is the mean
z is the confidence level value
s is the standard deviation
n is the population width
(a) The 90% confidence interval for the population mean
90% = 0.05
Z = 1.64
34 1.64 * 8 / √50
34 1.855
32.145 < < 35.855
(b) The 99% confidence interval for the population mean
99% = 0.005
Z=2.57
34 2.57 * 8 / √50
34 2.907
31.093 < < 36.907
Therefore the confidence interval for population mean has been determined.
The complete question is
A simple random sample of 50 items from a population width =7 resulted in a sample mean of 35. If required, round your answers to two decimal places.
a. Provide a 90% confidence interval for the population mean
b. Provide a 99% confidence interval for the population mean
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