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Lunna [17]
2 years ago
10

A university knows from historical data that 25% of students in an introductory statistics class withdraw before completing the

class. Assume that 16 students have registered for the course. What is the probability that exactly 2 will withdraw?
Mathematics
1 answer:
Vikki [24]2 years ago
8 0

Answer:

13.4%

Step-by-step explanation:

Use binomial probability:

P = nCr p^r q^(n-r)

where n is the number of trials,

r is the number of successes,

p is the probability of success,

and q is the probability of failure (1-p).

Here, n = 16, r = 2, p = 0.25, and q = 0.75.

P = ₁₆C₂ (0.25)² (0.75)¹⁶⁻²

P = 120 (0.25)² (0.75)¹⁴

P = 0.134

There is a 13.4% probability that exactly 2 students will withdraw.

You might be interested in
A recent study found that 51 children who watched a commercial for Walker Crisps (potato chips) featuring a long-standing sports
hichkok12 [17]

Answer:

1

The claim is that the mean amount of Walker Crisps eaten was significantly higher for the children who watched the sports celebrity- endorsed Walker Crisps commercial

2

The kind of test to use is a t -test because a t -test is used to check if there is a difference between means of a population

3

t  =  3.054

4

The p-value  is   p-value  =  P(Z >  3.054) = 0.0011291

5

The conclusion is  

There is sufficient evidence to conclude that the mean amount of Walker Crisps eaten was significantly higher for the children who watched the sports celebrity- endorsed Walker Crisps commercial

The test statistics is  

Step-by-step explanation:

From the question we are told that

   The first sample size is  n_1  =  51

    The first sample  mean is \mu_1  =  36

    The second sample size is  n_2  =  41

    The second sample  size is  \mu_2  =  25

     The first standard deviation is  \sigma _1  =  21.4 \  g

    The second standard deviation is  \sigma _2  =  12.8 \  g

  The  level of significance is  \alpha =  0.05

The  null hypothesis is  H_o  :  \mu_1 = \mu_ 2

The  alternative hypothesis is  H_a :  \mu_1 > \mu_2

Generally the test statistics is mathematically represented as

    t  =  \frac{\= x_1 - \= x_2}{ \sqrt{ \frac{s_1^2}{n_1}  + \frac{s_2^2}{n_2}  } }

=>   t  =  \frac{ 36 - 25}{ \sqrt{ \frac{ 21.4^2}{51}  + \frac{ 12.8^2}{41}  } }

=> t  =  3.054

The  p-value is mathematically represented as

     p-value  =  P(Z >  3.054)

Generally from the z table  

             P(Z >  3.054) =  0.0011291

=>   p-value  =  P(Z >  3.054) = 0.0011291

From the values obtained  we see that p-value  < \alpha so  the null hypothesis is rejected

Thus the conclusion is  

  There is sufficient evidence to conclude that the mean amount of Walker Crisps eaten was significantly higher for the children who watched the sports celebrity- endorsed Walker Crisps commercial

5 0
2 years ago
Convert decimal +49 and +29 to binary, using the signed‐2’s‐complement representation and enough digits to accommodate the numbe
Serhud [2]

To solve this problem, let us first find for the binary equivalents of the numbers. They are:

Decimal --> Binary

+ 29 --> 00011101

+ 49 --> 00110001

- 29 --> 11100011

- 49 --> 11001111

Now we apply the normal binary arithmetic to these converted numbers:

(+ 29) + (- 49) ---> 00011101 + 11001111 = 11101100 ---> - 20 (TRUE)

(- 29) + (+ 49) ---> 11100011 + 00110001 = 00010100 ---> + 20 (TRUE)

(- 29) + (- 49) ---> 11100011 + 11001111 = 10110010 ---> - 78 (TRUE)

5 0
2 years ago
A group of 75 math students were asked whether they like algebra and whether they like geometry. A total of 45 students like alg
Yuki888 [10]

Answer:

Step-by-step explanation:

Let x be the number of students that like both algebra and geometry. Then:

1. 45-x is the number of students that like only algebra;

2. 53-x is the number of students that like only geometry.

You know that 6 students do not like any subject at all and there are 75 students in total.

If you add the number of students that like both subjects, the number of students that like only one subject and the number of students that do not like any subject, you get 75.

Therefore,

x+45-x+53-x+6=75.

Solve this equation:

104-x=75,\\\\x=104-75,\\\\x=29.

You get that:

29 students like both subjects;

45-29=16 students like only algebra;

53-29=24 students like only geometry;

24+6=30 students do not like algebra;

16+6=22 students do not like geometry.

a = 29, b = 16, c = 24, d = 30, e = 22

<h3>The correct choice is D.</h3>

5 0
2 years ago
Read 2 more answers
What is the value of x when y=10 in the equation y=4x-2
Viefleur [7K]
Substitute y for 10 in the equation and solve using these steps:
10=4x+2
1) Minus 2 from both sides
8=4x
2) Divide both sides by 4 (the coefficient of x)
2=x
The value of x is 2 and we can check this by substituting it back into the original equation, 10 = (4×2)+2.
3 0
2 years ago
The probability of a train arriving on time and leaving on time is 0.8. The probability that the train arrives on time and leave
kkurt [141]

Answer:

<u>0.9524</u>

Step-by-step explanation:

<em>Note enough information is given in this problem. I will do a similar problem like this. The problem is:</em>

<em>The Probability of a train arriving on time and leaving on time is 0.8.The probability of the same train arriving on time is 0.84. The probability of the same train leaving on time is 0.86.Given the train arrived on time, what is the probability it will leave on time?</em>

<em />

<u>Solution:</u>

This is conditional probability.

Given:

  • Probability train arrive on time and leave on time = 0.8
  • Probability train arrive on time = 0.84
  • Probability train leave on time = 0.86

Now, according to conditional probability formula, we can write:

P(Leave \ on \  time | arrive \  on \ time) = P(arrive ∩ leave) / P(arrive)

Arrive ∩ leave means probability of arriving AND leaving on time, that is given as "0.8"

and

P(arrive) means probability arriving on time given as 0.84, so:

0.8/0.84 = <u>0.9524</u>

<u></u>

<u>This is the answer.</u>

5 0
2 years ago
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