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Alisiya [41]
3 years ago
8

The score of golfers for a particular course follows a normal distribution that has a mean of 73 and a standard deviation of 3.

Suppose a golfer played the course today. Find the probability that her score is at least 74. 0.4772
Mathematics
1 answer:
Artemon [7]3 years ago
3 0

Answer:

P(X \geq 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(\mu=73,\sigma^{2}=3^{2})

The z score probability distribution is given by;

           Z = \frac{X-\mu}{\sigma} ~ N(0,1)

where, \mu = population mean = 73

           \sigma = standard deviation = 3

So, the probability that the score of golfer is at least 74 is given by = P(X \geq 74)

 P(X \geq 74) = P( \frac{X-\mu}{\sigma} \geq \frac{74-73}{3} ) = P(Z \geq 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 .

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