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Alcohol, threats, and electric shocks. A group of Florida State University psychologists examined the effects of alcohol on the reactions of people to a threat (Journal of Abnormal Psychology, Vol. 107, 1998). After obtaining a specified blood alcohol level, the psychologists placed experimental subjects in a room and threatened them with electric shocks. Using sophisticated equipment to monitor the subjects’ eye movements, the psychologists recorded the startle response (measured in milliseconds) of each subject. The mean and standard deviation of the startle responses were 37.9 and 12.4, respectively. Assume that the startle response x for a person with the specified blood alcohol level is approximately normally distributed. Find the probability that x is between 40 and 50 milliseconds. Find the probability that x is less than 30 milliseconds. Give an interval for x centered around 37.9 milliseconds so that the probability that x falls in the interval is 0.95. Ten percent of the experimental subjects have startle responses above what value?

Answer :

To solve this problem, we will make use of the properties of the normal distribution. Here’s how we can tackle each part of the question:


  1. Probability that [tex]x[/tex] is between 40 and 50 milliseconds:


    • Given that the startle response [tex]x[/tex] is normally distributed with mean [tex]\mu = 37.9[/tex] and standard deviation [tex]\sigma = 12.4[/tex].

    • We standardize the values 40 and 50 using the formula for the z-score: \
      [tex]z = \frac{x - \mu}{\sigma}[/tex]

    • For [tex]x = 40[/tex]:
      [tex]z_{40} = \frac{40 - 37.9}{12.4} \approx 0.169[/tex]

    • For [tex]x = 50[/tex]:
      [tex]z_{50} = \frac{50 - 37.9}{12.4} \approx 0.976[/tex]

    • We now use standard normal distribution tables or software to find the probabilities:
      [tex]P(0.169 \leq Z \leq 0.976) \approx P(Z \leq 0.976) - P(Z \leq 0.169) \approx 0.834 - 0.567 = 0.267[/tex]

    • The probability that [tex]x[/tex] is between 40 and 50 milliseconds is approximately 0.267.



  2. Probability that [tex]x[/tex] is less than 30 milliseconds:


    • Calculate the z-score for [tex]x = 30[/tex]:
      [tex]z_{30} = \frac{30 - 37.9}{12.4} \approx -0.637[/tex]

    • Using the z-table,
      [tex]P(Z \leq -0.637) \approx 0.261[/tex]

    • So, the probability that [tex]x[/tex] is less than 30 milliseconds is approximately 0.261.



  3. Interval for [tex]x[/tex] centered around 37.9 milliseconds with 0.95 probability:


    • For a 95% confidence interval, we find the z-values corresponding to the central 95% of the normal distribution, which are approximately [tex]z = \pm 1.96[/tex].

    • Calculate the interval:
      [tex]x = \mu \pm z \times \sigma[/tex]
      [tex]x = 37.9 \pm 1.96 \times 12.4[/tex]
      [tex]x = 37.9 \pm 24.30[/tex]
      [tex]\text{Interval}: (13.6, 62.2)[/tex]

    • Thus, the interval centered around 37.9 milliseconds so that the probability [tex]x[/tex] falls within this interval is 0.95 is approximately (13.6, 62.2).



  4. Ten percent of the experimental subjects have startle responses above what value?


    • This requires finding the 90th percentile (i.e., the z-score where [tex]P(Z < z) = 0.9[/tex]).

    • The z-score for the 90th percentile is approximately 1.28.

    • Compute the corresponding [tex]x[/tex] value:
      [tex]x = \mu + z \times \sigma[/tex]
      [tex]x = 37.9 + 1.28 \times 12.4[/tex]
      [tex]x = 37.9 + 15.872 \approx 53.772[/tex]

    • Therefore, ten percent of the experimental subjects have startle responses above approximately 53.772 milliseconds.



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