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algol [13]
4 years ago
14

Mr. Green teaches mathematics and his class recently finished a unit on statistics. The student scores on this unit are: 40 47 5

0 50 50 54 56 56 60 60 62 62 62 63 65 70 70 72 76 77 80 85 85 95 Perform a statistical analysis on this data set. a) Find the measures of central tendency and the range of the data.
Mathematics
1 answer:
Harrizon [31]4 years ago
5 0

Answer:

Mean = 64.46, Median = 62 and Mode = Bi-modal (50 and 62)

Range of the data is 55.

Step-by-step explanation:

We are given that Mr. Green teaches mathematics and his class recently finished a unit on statistics.

<u>The student scores on this unit are:</u>  40, 47, 50, 50, 50, 54, 56, 56, 60, 60, 62, 62, 62, 63, 65, 70, 70, 72, 76, 77, 80, 85, 85, 95.

We know that Measures of Central Tendency are: Mean, Median and Mode.

  • Mean is calculated as;

                   Mean  =  \frac{\sum X}{n}

where  \sum X = Sum of all values in the data

               n = Number of observations = 24

So, Mean  =  \frac{40+ 47+ 50+ 50+ 50+ 54+ 56+ 56+ 60 +60+ 62+ 62+ 62+ 63+ 65+ 70+ 70+ 72+ 76+ 77+ 80+ 85+ 85+ 95}{24}

=  \frac{1547}{24}  =  64.46

So, mean of data si 64.46.

For calculating Median, we have to observe that the number of observations (n) is even or odd, i.e.;

  • If n is odd, then the formula for calculating median is given by;

                     Median  =  (\frac{n+1}{2})^{th} \text{ obs.}

  • If n is even, then the formula for calculating median is given by;

                     Median  =  \frac{(\frac{n}{2})^{th}\text{ obs.} +(\frac{n}{2}+1)^{th}\text{ obs.}   }{2}

Now here in our data, the number of observations is even, i.e. n = 24.

So, Median  =  \frac{(\frac{n}{2})^{th}\text{ obs.} +(\frac{n}{2}+1)^{th}\text{ obs.}   }{2}

                    =  \frac{(\frac{24}{2})^{th}\text{ obs.} +(\frac{24}{2}+1)^{th}\text{ obs.}   }{2}

                    =  \frac{(12)^{th}\text{ obs.} +(13)^{th}\text{ obs.}   }{2}

                    =  \frac{62 + 62  }{2}  =  \frac{124}{2}  =  62

Hence, the median of the data is 62.

  • A Mode is a value that appears maximum number of times in our data.

In our data, there are two values which appear maximum number of times, i.e. 50 and 62 as these both appear maximum 3 times in the data.

This means our data is Bi-modal with 50 and 62.

  • The Range is calculated as the difference between the highest and lowest value in the data.

                      Range  =  Highest value - Lowest value

                                   =  95 - 40 = 55

Hence, range of the data is 55.

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Answer:

a) Let the random variable X= "number of these tracks where SBIRS detects the object." in order to use the binomial probability distribution we need to satisfy some conditions:

1) Independence between the trials (satisfied)

2) A value of n fixed , for this case is 20 (satisfied)

3) Probability of success p =0.2 fixed (Satisfied)

So then we have all the conditions and we can assume that:

X \sim Bin(n =20, p=0.8)

b) X \sim Bin(n =20, p=0.8)

c) P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456

d) P(X \geq 15) = P(X=15)+ .....+P(X=20)

P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456

P(X=16)=(20C16)(0.8)^{16} (1-0.8)^{20-16}=0.218

P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205

P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137

P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.058

P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.012

P(X\geq 15)=0.804208

e) E(X) = np = 20*0.8 = 16

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

Part a

Let the random variable X= "number of these tracks where SBIRS detects the object." in order to use the binomial probability distribution we need to satisfy some conditions:

1) Independence between the trials (satisfied)

2) A value of n fixed , for this case is 20 (satisfied)

3) Probability of success p =0.2 fixed (Satisfied)

So then we have all the conditions and we can assume that:

X \sim Bin(n =20, p=0.8)

Part b

X \sim Bin(n =20, p=0.8)

Part c

For this case we just need to replace into the mass function and we got:

P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456

Part d

For this case we want this probability: P(X\geq 15)

And we can solve this using the complement rule:

P(X \geq 15) = P(X=15)+ .....+P(X=20)

P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456

P(X=16)=(20C16)(0.8)^{16} (1-0.8)^{20-16}=0.218

P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205

P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137

P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.058

P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.012

P(X\geq 15)=0.804208

Part e

The expected value is given by:

E(X) = np = 20*0.8 = 16

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