(x - 3)(x2 + 3x + 9)
1 answer:
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Answer: 1.25
Step-by-step explanation:
Given: A college-entrance exam is designed so that scores are normally distributed with a mean = 500 and a standard deviation
= 100.
A z-score measures how many standard deviations a given measurement deviates from the mean.
Let Y be a random variable that denotes the scores in the exam.
Formula for z-score =
Z-score =
⇒ Z-score =
⇒Z-score =1.25
Therefore , the required z-score = 1.25
Answer:
P(X 74) = 0.3707
Step-by-step explanation:
We are given that the score of golfers for a particular course follows a normal distribution that has a mean of 73 and a standard deviation of 3.
Let X = Score of golfers
So, X ~ N()
The z score probability distribution is given by;
Z = ~ N(0,1)
where, = population mean = 73
= standard deviation = 3
So, the probability that the score of golfer is at least 74 is given by = P(X 74)
P(X 74) = P(
) = P(Z
0.33) = 1 - P(Z < 0.33)
= 1 - 0.62930 = 0.3707
Therefore, the probability that the score of golfer is at least 74 is 0.3707 .