Question

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

Answer 1

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 .