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katrin2010 [14]
3 years ago
14

In general, after a group makes a reservation at a restaurant, they do not show up 20% of the time. If a restaurant has 8 reserv

ations for the evening, what is the probability that exactly two do not show up?
Mathematics
1 answer:
sashaice [31]3 years ago
5 0

Answer:

P(more than 2 don't show up)=0.2031

Step-by-step explanation:

Multiplication rule for independent events:

If two events A and B are independent then the multiplication rule state that:

P(A and B)= P(A) x P(B)

Find the probability that more than two do not show up:

The probability that the reservations at the restaurant do not show up is 0.20 (=20%) and there were 8 reservations.

P(more than 2 don't show up)= 1- P(less than or equal to 2 don't show up)

= 1-( P(none show up) +P(one don't show up)+P(two don't show up)

=1- ((0.8⁸) +₈ C₁ (0.2¹)x(0.8⁷)  +₈ C₂ (0.2²)x(0.8⁶)

=1- (0.1678 +0.3355+0.2936)= 1-0.7969

P(more than 2 don't show up)=0.2031

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There are 10 employees in a particular division of a company. Their salaries have a mean of 570,000, a median of $55,000,and a s
harkovskaia [24]

Answer:

a) $160,000

b) $55,000

c) $332264.804

Step-by-step explanation:

We are given that there are 10 employees in a particular division of a company and their salaries have a mean of $70,000, a median of $55,000, and a standard deviation of $20,000.

And also the largest number on the list is $100,000 but By accident, this number is changed to $1,000,000.

a) Value of mean after the change in value is given by;

     Original Mean = $70,000

       \frac{\sum X}{n} = $70,000  ⇒ \sum X = 70,000 * 10 = $700,000

   New \sum X after change = $700,000 - $100,000 + $1,000,000 = $1600000

  Therefore, New mean = \frac{1600000}{10} = $160,000 .

b) Median will not get affected as median is the middle most value in the data set and since $1,000,000 is considered to be an outlier so median remain unchanged at $55,000 .

c) Original Variance = 20000^{2} i.e.  20000^{2} = \frac{\sum x^{2} - n*xbar }{n -1}

    Original \sum x^{2} = (20000^{2} * (10-1)) + (10 * 70,000) = $3,600,700,000

    New \sum x^{2} = $3,600,700,000 - 100,000^{2} + 1,000,000^{2} = 9.936007 * 10^{11}  

    New Variance = \frac{new\sum x^{2} - n*new xbar }{n -1} = \frac{9.936007 *10^{11}  - 10*160000 }{10 -1} = 1.103999 * 10^{11}    Therefore, standard deviation after change = \sqrt{1.103999 * 10^{11} } = $332264.804 .

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3 years ago
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