Download 3) Here it is give that the value of z is such that 48% of

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3) Here it is give that the value of z is such that 48% of the distribution lies between it
and the mean.
That is, either P[z < Z < 0] = 0.48 or P[0 < Z < z] = 0.48
Or equivalently, P[Z >z] = 0.98 or P[Z < z] = 0.98 (Note that P[Z>0]=P[Z<0] = 0.5)
That is, either z = -2.0537 or 2.0537
[The value of z such that P[Z < z] = 0.98 can be calculated using the Excel formula
=NORMSINV(0.98)]
4) Let X denote the average annual salary for a worker in the United States. Here it is
given that X follows a Normal distribution with mean $41,000 and standard deviation
$7,000. Then Z = (X – 41000)/7000, follows a Standard Normal distribution.
A) The percentage of Americans earn below $27,000 is given by,
P[X < $27,000] = P[(X – 41000)/7000 < (27000 – 41000)/7000]
= P[Z < -2]
= 0.0228
= 2.28%
Note that the probability P[Z < -2] can be calculated using the Excel formula
=NORMSDIST(-2)
B) The percentage of Americans earn above $43,000 is given by,
P[X > $43,000] = P[(X – 41000)/7000 > (42000 – 41000)/7000]
= P[Z > 0.1429]
= 1 - P[Z < 0.1429]
= 1 - 0.5568 = 0.4432
= 44.32%
Note that the probability P[Z < 0.1429] can be calculated using the Excel formula
=NORMSDIST(0.1429)
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