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s2008m [1.1K]
3 years ago
13

according to a recent article about individuals who have credit cards, the mean number of cards per person with credit cards is

4. To test this results a random survey of 60 individuals who have credit cards were conducted. The survey only includes the number of credit cards per participant

Mathematics
1 answer:
Afina-wow [57]3 years ago
5 0

Answer:

The variable of interest is NUMBER OF CREDIT CARDS. it is a QUANTITATIVE variable.

u=4

u<4

use a t1-84 go to stat, edit, input the table set into L1, back to stat, tests, t-test, data, u=4, list:L1, u<, calculate, enter that will give you a t0=-.62

p-value is in the range P-value>0.10

The P-value is GREATER than level of significance. DO NOT REJECT the null hypothesis. there IS NOT sufficient evidence to claim that the mean number of credit cards is LESS THAN 4.

Step-by-step explanation:

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Problems 27: the number X is the number of car which passed through the
svet-max [94.6K]

Answer:

a) The value of a= 0.59

b)  The probability that there are at-least 3 cars passing through the stop sign

P(x>3)  = 0.03

c)

The Expected value of X  = 0.62

d)

 The variance of X is   σ² = 0.9556

e)

The standard deviation of X

                                    σ = 0.9775

Step-by-step explanation:

<u><em></em></u>

Given data

X       :      0         1          2            3        4          5

P(x)   :    a       0.30     0.05    0.03    0.02     0.01

a)

      ∑ p(x_{i} ) = 1

 a + 0.30+ 0.05+ 0.03+ 0.02+0.01 = 1

 a  + 0.41 = 1

     a  = 1 - 0.41

   <em>  a  = 0.59</em>

b)

<em>  The probability that there are at-least 3 cars passing through the stop sign</em>

<em>P(x >3)  = P( x=4) + P( x=5)</em>

           =   0.02 +0.01

          = 0.03

c)

X       :      0           1          2            3        4          5

P(x)   :      0.59      0.30     0.05    0.03    0.02     0.01

The Expected value of X

E(X) = ∑ x P(X= x)

      =  0 + 1 ×0.30 + 2×0.05 + 3×0.03 + 4×0.02 + 5×0.01

     =   0.30 + 0.1 + 0.09 +0.08 +0.05

    =  0.62

<em>The Expected value of X </em>

<em>                           E(X)  = 0.62</em>

<em>d) </em>

   The variance of the discrete distribution

       σ² =   ∑ x²p(x) -μ²

      σ² =   0 + 1² ×0.30 + 2² ×0.05 + 3² ×0.03 + 4² ×0.02 + 5²× 0.01 - (0.62)²

          =   1.34 - 0.3844

         = 0.9556

      σ² = √0.9556

<em>e) The standard deviation of the  discrete distribution</em>

<em>      σ = 0.9775</em>

<em></em>

4 0
3 years ago
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