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3 years ago

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According to a study conducted in one city, 35% of adults in the city have credit card debts more than $2.000. A simple random s

ample of n=250 adults is obtained from the city. Describe the sampling distribution of P^, the sample proportion of adults who have credit card debts of more than $2000.(Round to three decimal places when necessary.)Select from one of the 4 answers belowA. approximately- normal; \mu p=0.35, \sigma p=0.030B. approximately- normal; \mu p=0.35, \sigma p=0.001C. exactly- normal; \mu p=0.35, \sigma p=0.030D.Binomial; \mu p=87.5, \sigma p=7.542

1 answer:

BabaBlast [244]3 years ago

4 0

Answer:

Binomial; \mu p=87.5, \sigma p=7.542

Step-by-step explanation:

a distribution is said be a binomial distribution iff

The probability of success of that event( let it be p) is same for every trial
each trial should have 2 outcome : p or (1-p) i.e, success or failure only.
there are fixed number of trials (n)
the trials are independent

here, the trials are obviously independent ( because, one person's debt doesn't influence the other person's)

here n=250

the probability of success(0.35) is same for every trial

(35/100=0.35 is the required p here)

$\mu_{p} =n*p=250*\frac{35}{100} =250*0.35=87.5$

[since, the formula for $\mu _{p} =n*p$ ]

$\sigma _{p} =\sqrt{n*(p)*(1-p)} = \sqrt{250*0.35*(1-0.35)} = 7.542 ( approximately)$

[since, the formula for [tex]\sigma _{p} =\sqrt{n*(p)*(1-p)}

therefore, it is Binomial; \mu p=87.5, \sigma p=7.542

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