The sum of 15 and the product of five and V
1 answer:
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34% of the scores lie between 433 and 523.
Solution:
Given data:
Mean (μ) = 433
Standard deviation (σ) = 90
<u>Empirical rule to determine the percent:</u>
(1) About 68% of all the values lie within 1 standard deviation of the mean.
(2) About 95% of all the values lie within 2 standard deviations of the mean.
(3) About 99.7% of all the values lie within 3 standard deviations of the mean.
Z lies between o and 1.
P(433 < x < 523) = P(0 < Z < 1)
μ = 433 and μ + σ = 433 + 90 = 523
Using empirical rule, about 68% of all the values lie within 1 standard deviation of the mean.
i. e.
Here μ to μ + σ =
Hence 34% of the scores lie between 433 and 523.
Answer: a.) $50188 to $57812
Step-by-step explanation: <u>Confidence</u> <u>Interval</u> (CI) is an interval of values in which we are confident the true mean is in.
The interval is calculated as
x ±
a. For a 95% CI, z-value is 1.96.
Solving:
54,000 ±
54,000 ±
54,000 ± 1.96*1732.102
54,000 ± 3395
This means the interval is
50605 < μ < 57395
<u>With a 95% confidence interval, the mean starting salary of college graduates is between 50605 and 57395 or </u><u>from 50188 to 57812$.</u>
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b. The mean starting salary for college students in 2017 is $50,516, which is in the confidence interval. Therefore, since we 95% sure the real mean is between 50188 and 57812, there was no significant change since 2017.