The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three. What the probability that a

Question

The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three. What the probability that a randomly selected golfer scored less than 65?

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2 weeks 2021-11-15T13:05:37+00:00 1 Answer 0 views 0

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    2021-11-15T13:06:40+00:00

    Answer:

    P(X<65)=P(\frac{X-\mu}{\sigma}<\frac{65-\mu}{\sigma})=P(Z<\frac{65-68}{3})=P(Z<-1)

    And we can find this probability using the normal standard table or excel and we got:

    P(Z<-1)=0.159

    Step-by-step explanation:

    Previous concepts

    Normal distribution, is a “probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean”.

    The Z-score is “a numerical measurement used in statistics of a value’s relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean”.  

    Solution to the problem

    Let X the random variable that represent the golf scores of a population, and for this case we know the distribution for X is given by:

    X \sim N(68,3)  

    Where \mu=68 and \sigma=3

    We are interested on this probability

    P(X<65)

    And the best way to solve this problem is using the normal standard distribution and the z score given by:

    z=\frac{x-\mu}{\sigma}

    If we apply this formula to our probability we got this:

    P(X<65)=P(\frac{X-\mu}{\sigma}<\frac{65-\mu}{\sigma})=P(Z<\frac{65-68}{3})=P(Z<-1)

    And we can find this probability using the normal standard table or excel and we got:

    P(Z<-1)=0.159

    We can use the following excel code to find it: “=NORM.DIST(-1,0,1,TRUE)”

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