Solution :
Given that ,
mean = = 87
standard deviation = = 36
a)
C) The distribution is approximately normal
b)
n = 81
= = 87 and
= / n = 36 / 81 = 4
b)
P( > 94.2) = 1 - P( < 94.2)
= 1 - P(( - ) / < (94.2 - 87) / 4)
= 1 - P(z < 1.8)
= 1 - 0.9641 Using standard normal table.
= 0.0359
Probability = 0.0359
c)
P( 77.2) = P(( - ) / (77.2 - 87) / 4)
= P(z < -2.45) Using standard normal table.
Probability = 0.0071
d)
P( 82.4 < < 95.8) = P((82.4 - 87) /4 <( - ) / < (95.8 - 87) / 4))
= P(-1.15 < Z < 2.2)
= P(Z < 2.2) - P(Z < -1.15) Using standard normal table,
= 0.9861 - 0.1251
= 0.8610
Probability = 0.8610
A simple random sample of size n = 81 is obtained from a population with u...
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