A chocolate chip cookies recipe calls for 4 and 1/4 cups of chocolate chips to make 10 dozen cookies Based on this recipe, how m
2 answers:
You might be interested in
Let X be the national sat score. X follows normal distribution with mean μ =1028, standard deviation σ = 92
The 90th percentile score is nothing but the x value for which area below x is 90%.
To find 90th percentile we will find find z score such that probability below z is 0.9
P(Z <z) = 0.9
Using excel function to find z score corresponding to probability 0.9 is
z = NORM.S.INV(0.9) = 1.28
z =1.28
Now convert z score into x value using the formula
x = z *σ + μ
x = 1.28 * 92 + 1028
x = 1145.76
The 90th percentile score value is 1145.76
The probability that randomly selected score exceeds 1200 is
P(X > 1200)
Z score corresponding to x=1200 is
z =
z =
z = 1.8695 ~ 1.87
P(Z > 1.87 ) = 1 - P(Z < 1.87)
Using z-score table to find probability z < 1.87
P(Z < 1.87) = 0.9693
P(Z > 1.87) = 1 - 0.9693
P(Z > 1.87) = 0.0307
The probability that a randomly selected score exceeds 1200 is 0.0307