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yarga [219]
1 year ago
13

In 1984, an intruder broke into the home of Jennifer Thompson, a 22-year-old student, and sexually assaulted her. In the initial

stages of the subsequent investigation, the police showed Jennifer a photo array of men matching the description she gave of the intruder. She chose a photo of Ronald Cotton and identified him as the assailant. At a later point in the investigation, Jennifer identified Ronald Cotton as her attacker a second time by choosing him from a lineup. Mr. Cotton spent 11 years in prison before DNA testing proved that someone else, a man named Bobby Poole, had actually attacked Jennifer. Although correct procedures were followed in the investigation, a combination of several factors led to Jennifer's confident and persuasive testimony during Ronald Cotton's trial. Which of the following presents a scenario where positive feedback could have impacted Jennifer's identification of her assailant? a. Jennifer was given only ten minutes to identify her attacker from the lineup. b. Jennifer was shown several photos simultaneously rather than a sequence of photos.c. Jennifer was not informed that her attacker might not be in the lineup. d. Officers told Jennifer that the man she identified in the photo array had been involved in other breaking-and-entering crimes.
Social Studies
1 answer:
Hunter-Best [27]1 year ago
8 0

Answer:

d. Officers told Jennifer that the man she identified in the photo array had been involved in other breaking-and-entering crimes.

Explanation:

<em>Jennifer identifcation of the man at the initial stage he was arranged could have being filled with doubt but during the second arrangement of those people, she could have been biased in sticking to her first choice.</em><em>This could be as a result of learning from the investigating officer that, the person she identified earlier has been involved in other breakings-and-entering crimes.</em>

<em>This information would definitely have influenced her judgement during the second parade of those people</em>. Had it been such information was actually withheld from her, there is a great chance of her backtracking and withdrawing from her first choice and probably taking a judgement free from bias.

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A section of highway is posted with a 55 mph speed limit. In fact, speeds of cars traveling on this section are normally distrib
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Answer:

z=1.28

And if we solve for a we got

a=58 +1.28*5=64.4

So the value of height that separates the bottom 90% of data from the top 10% is 64.4.  

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the velocity of a population, and for this case we know the distribution for X is given by:

X \sim N(58,5)  

Where \mu=58 and \sigma=5

For this part we want to find a value a, such that we satisfy this condition:

P(X>a)=0.10   (a)

P(X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.90 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=1.28

And if we solve for a we got

a=58 +1.28*5=64.4

So the value of height that separates the bottom 90% of data from the top 10% is 64.4.  

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