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Varvara68 [4.7K]
2 years ago
9

The times of the runners in a marathon are normally distributed, with a mean of 3 hours and 50 minutes and a standard deviation

of 30 minutes. what is the probability that a randomly selected runner has a time less than or equal to 3 hours and 20 minutes? use the portion of the standard normal table below to help answer the question.
Mathematics
1 answer:
serious [3.7K]2 years ago
4 0

The probability that a randomly selected runner has a time less than or equal to 3 hours and 20 minutes is 0.16 or 16%.

<h3>What is normally distributed data?</h3>

Normally distributed data is the distribution of probability which is symmetric about the mean.

The mean of the data is the average value of the given data.

The standard deviation of the data is the half of the difference of the highest value and mean of the data set.

The times of the runners in a marathon are normally distributed, with

  • Mean of 3 hours and 50 minutes
  • Standard deviation of 30 minutes.

Refere the probabiliity table attached below. The probability of Z being inside the 1 Standard daviation of mean is 0.84.

The probability of runner selected with time less than or equal to 3 hours and 20 minutes,

P=1-0.84\\P=0.16

Thus, the probability that a randomly selected runner has a time less than or equal to 3 hours and 20 minutes is 0.16 or 16%.

Learn more about the normally distributed data here;

brainly.com/question/6587992

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Assume that different groups of couples use the XSORT method of gender selection and each couple gives birth to one baby. The XS
olganol [36]

Answer:

Mean and Standard deviation for the numbers of girls in groups of 36 births are 18 and 3 respectively.

Step-by-step explanation:

We are given that he X SORT method is designed to increase the likelihood that a baby will be a girl, but assume that the method has no effect, so the probability of a girl is 0.5.

Now consider a group consisting of 36 couples.

The above situation can be represented through binomial distribution;

P(X=r)=\binom{n}{r} \times p^{r} \times (1-p)^{n-r} ;x=0,1,2,3,.....

where, n = number of trials (samples) taken = 36 couples

            r = number of success

            p = probability of success which in our question is probability

                   of a girl, i.e.; p = 0.5

<em><u>Let X = Numbers of girls in groups of 36 births </u></em>

So, X ~ Binom(n = 36, p = 0.5)

Now, mean for the numbers of girls in groups of 36 births is given by;

         <u>Mean</u>, E(X) = n \times p = 36 \times 0.5 = 18

Also, standard deviation for the numbers of girls in groups of 36 births is given by;

        <u>Standard deviation</u>, S.D.(X) =  \sqrt{n\times p\times (1-p)}

                                                      =  \sqrt{36\times 0.5\times (1-0.5)}

                                                      =  3

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3 years ago
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