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So here's the question: Maria is shopping for wireless routers and is overwhelmed by the number of options. In order to get a feel for the average price, she takes a random sample of $75$ routers. The average price for this sample is $\$76$ and the standard deviation is $\$30$.

A consumer information website claims that the average price of all routers is $\$80$. If we assume that the population standard deviation of prices is $\$30$ (i.e., if we assume Maria's sample standard deviation is a good estimate of $\sigma$), what is the probability that a random sample of $75$ routers would have an average price of less than or equal to $\$76$? Give your answer to three decimal places.

How do I solve this problem? I'm completely stuck and would really appreciate any help.

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  • $\begingroup$ Thank you for the edit lulu! $\endgroup$
    – Jacob
    Oct 18, 2015 at 20:36

2 Answers 2

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Here are some hints to help you structure your thinking:

  1. If $\sigma_{X_i} = 30$ what is the standard deviation of $\frac{1}{75}\sum_{1}^{75} X_i$? (Let's call this $\sigma_{\bar X}$)
  2. What is the $Z$ score of the difference $D=76-80$?
  3. What is the probability that a standard normal random variable will take on values less than $Z$?

To put OP out of his misery...

We are told that the router prices are normally distributed with a standard deviation of 30 and a mean of 80. What this problem is asking you to do is determine how likely it is that Maria's sample of 75 routers came from a normally distributed population with mean 80, standard deviation 30.

The first thing to do is calculate the standard deviation of the sample average of 75 iid observations from this population. As you correctly calculated, that would be $\frac{30}{\sqrt{75}} \approx 3.46$. So, 95% of the time, a sample average of 75 iid observations from a normally distributed population of mean 80, standard deviation 30 will fall i the interval $80 \pm 2\times 3.46$.

Maria's sample average is 76. The question is asking you for the probability of getting less than or equal to 76 for the sample average if it came from a population with mean 80 and standard deviation 30.

How many standard deviations below the mean is the sample average of $76$?

We already calculated that the standard deviation of a 75 observation mean to be 3.46, so 76 is $\frac{76-80}{3.46}\approx -1.15$ standard deviations below the expected sample average.

How likely is this? We simply need to look up the above value (-1.15) in a standard normal table (you should have been taught how to do this already) or you can use excels NORMSDIST(-1.15) to get a lower tailed probability. This will give you 0.12 or $12\%$

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  • $\begingroup$ Would I simply run norm.dist(75, 76, 30, true) in excel? $\endgroup$
    – Jacob
    Oct 18, 2015 at 20:43
  • $\begingroup$ @Jacob Try to work though the above steps. It will require more than one step. $\endgroup$
    – user237392
    Oct 18, 2015 at 20:44
  • $\begingroup$ I got 1 to be 3.464. For 2 I got 4. 3 I don't understand. $\endgroup$
    – Jacob
    Oct 18, 2015 at 20:46
  • $\begingroup$ @Jacob 1 is correct. 2 is not correct. Do you know what a Z-score is? Its the difference divided by the standard deviation. For 3, have you used standard normal tables or Excels norm.s.dist? $\endgroup$
    – user237392
    Oct 18, 2015 at 20:49
  • $\begingroup$ For 2, I got 0.13. For 3 I got 0.05. $\endgroup$
    – Jacob
    Oct 18, 2015 at 20:52
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8.) A population has to pass a quality test based on data from past runs there is a probability that 93.9% percent will pass and 4.1% will fail, there are 176 units in the population. What is the standard deviation for a sample of 50 and 75 units, what is the probability of having 2 failed units in a sample of 50 and 75.

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