Suppose that in one region of the country, the mean amount of credit card debt per household in...

Suppose that in one region of the country, the mean amount of credit card debt per household in...

Question:

Suppose that in one region of the country, the mean amount of credit card debt per household in households having credit card debt is $15,250 with standard deviation $7,125. The probability that the mean amount of credit card debt in a sample of 1600 such households will be within $300 of the population mean is roughly ...

Normal distribution

The normal distribution helps to calculate the probability. The probability of continuous distribution is estimated for left tailed, right-tailed, and between the value of a random variable.

 

Answer and Explanation:

Given information

  • Mean: 15250
  • Standard deviation: 7125
  • Sample size: 1600

The probability that the mean amount of credit card debt in a sample of 1600 such households will be within 300 is calculated as follows.

P(μ−300<X¯<μ+300)=P(μ−300−μ71251600<X¯−μσn<μ+300−μ71251600)=P(−1.68<Z<1.68)=P(Z<1.68)−P(Z<−1.68)(From standard normal table)=0.9535−0.0464=0.9071">P(μ?300<ˉX<μ+300)=P?? ? ? ??μ?300?μ7125√1600<ˉX?μσ√n<μ+300?μ7125√1600?? ? ? ??=P(?1.68<Z<1.68)=P(Z<1.68)?P(Z<?1.68)(From standard normal table)=0.9535?0.0464=0.9071P(μ?300

Therefore, the required probability is 0.9071

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