Write the percent and fraction equivalents. 1.07 please help!
1 answer:
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Answer:
2.28% probability that a person selected at random will have an IQ of 110 or greater
Step-by-step explanation:
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.
In this problem, we have that:
What is the probability that a person selected at random will have an IQ of 110 or greater?
This is 1 subtracted by the pvalue of Z when X = 110. So
has a pvalue of 0.9772
1 - 0.9772 = 0.0228
2.28% probability that a person selected at random will have an IQ of 110 or greater
Answer:
Therefore the 95% confidence interval is (25,707.480 < E < 26,744.920)
Step-by-step explanation:
n = 77
mean u = 26,226.2 bushels per acre
standard deviation s = 2,322.32
let E = true mean
let A = test statistic
Find 95% Confidence Interval
so
let A = (u - E) * ( / s) be the test statistic
we want P( average_l < A < average_u ) = 95%
look for lower 2.5% and the upper 97.5% Because I think this is a 2-tail test
average_l = -1.96 which corresponds to the 2.5%
average_u = 1.96
P( -1.96 < A < 1.96) = 95%
P( -1.96 < (u - E) * ( / s) < 1.96) = 95%
Solve for the true mean E ok
E < u + 1.96* (s / )
from -1.96 < (u - E) * ( / s)
E < 26,226.2 + 1.96*( 2,322.32 / )
E < 26,226.2 + 1.96*( 2,322.32 / )
E < 26,226.2 + 518.7197348105429466
upper bound is 26,744.9197
or
u - 1.96* (s / ) < E
26,226.2 - 518.7197348105429466 < E
25,707.48026519 < E
lower bound is 25,707.48026519
Therefore the 95% confidence interval is (25,707.480 < E < 26,744.920)