8)
The test scores are normally distributed with a mean of 50 and a standart deviation of 5. If P (X > X1) = 0.1562.
X1 = ?
Given that,
mean = = 50
standard deviation = =5
Using standard normal table,
P(Z > z) = 0.1562
= 1 - P(Z < z) = 0.1562
= P(Z < z ) = 1 - 0.1562
= P(Z < z ) = 0.8438
z = 1.01 (using standard normal (Z) table )
Using z-score formula
x = z * +
x= 1.01 *5+50
x= 55.05
X1 = 55.05
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