Simplify 5 − 2(2.5 − 6.7)^2
1 answer:
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Answer:
The correct option is (b).
Step-by-step explanation:
If X N (µ, σ²), then
, is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, Z
N (0, 1).
The distribution of these z-variate is known as the standard normal distribution.
The mean and standard deviation of the active minutes of students is:
<em>μ</em> = 60 minutes
<em>σ </em> = 12 minutes
Compute the <em>z</em>-score for the student being active 48 minutes as follows:
Thus, the <em>z</em>-score for the student being active 48 minutes is -1.0.
The correct option is (b).
Answer:
0.8973
Step-by-step explanation:
Relevant data provided in the question as per the question below:
Free throw shooting percentage = 0.906
Free throws = 6
At least = 5
Based on the above information, the probability is
Let us assume the X signifies the number of free throws
So, Then X ≈ Bin (n = 6, p = 0.906)
Now
The Required probability = P(X ≥ 5) = P(X = 5) + P(X = 6)
= 0.8973