Given:
mean, μ = 11 lb
Std. deviation, σ = 2.1
Let x = the weight that separates the top 5% of the crop.
The z-score is
z = (x - μ)/σ
= (x - 11)/2.1
From standard z-table for normal distribution,
P(z=1.645) = 0.95
Therefore
(x - 11)/2.1 = 1.645
x - 11 = 1.645/2.1 = 0.7833
x = 11.783
Answer: 11.8 lb (nearest tenth)
Answer:
y = -6/5 -3
Step-by-step explanation:
Point-Slope Form: (y - y_0)=m(x - x_0)
Answer:
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Answer:
a
Step-by-step explanation:
Answer:
81.86%
Step-by-step explanation:
We have been given that final exam scores are normally distributed with a mean of 74 and a standard deviation of 6.
First of all we will find z-score using z-score formula.
Now let us find z-score for 86.
Now we will find P(-1<Z) which is probability that a random score would be greater than 68. We will find P(2>Z) which is probability that a random score would be less than 86.
Using normal distribution table we will get,
We will use formula 
to find the probability to find that a normal variable lies between two values.
Upon substituting our given values in above formula we will get,
Upon converting 0.81859 to percentage we will get
Therefore, 81.86% of final exam score will be between 68 and 86.
