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Andrew [12]
3 years ago
8

In the flexed arm​ condition, what is the probability that a consumer has a choice score of

Mathematics
1 answer:
Flauer [41]3 years ago
6 0
<span>A study was conducted to determine if consumers are more likely to choose a vice product (e.g., a candy bar) when their arm is flexed (as when carrying a shopping basket) than when their arm is extended(as when pushing a shopping cart). The study measured choice scores (on a scale of 0 to 100, where higher scores indicate a greater preference for vice options) for consumer shopping under each of the two conditions. The average choice score for consumers with a flexed arm was 60, while the average for consumers with an extended arm was 43. For both conditions, assume that the standard deviation of the choice scores is 6. Also assume that both distributions are approximately normally distributed. Complete part a. a. In the flexed arm condition, what is the probability that a consumer has a choice score of 56 or greater? what is the probability that a consumer in the extended arm condition has a choice score of 56 or greater?</span>
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topjm [15]

Option C

<u>SOLUTION:</u>

We need to find the value of B - CF

First find the value CF:

CF=\left[\begin{array}{ccc}12&0&1.5\\1&-6&7\\\end{array}\right]  \left[\begin{array}{ccc}-2&0\\0&8\\2&1\end{array}\right]

CF=\left[\begin{array}{ccc}12(-2)+0 *0+1.5*2&12*0+0.8+1.5*1\\1*(-2)+(-6)*0+7.2&1*0+(-6)*8+7.1\\\end{array}\right]

CF=\left[\begin{array}{ccc}-21&1.5\\12&-41\\\end{array}\right]

Now find value of B - CF:

B-CF=\left[\begin{array}{ccc}2&8\\6&3\\\end{array}\right] -\left[\begin{array}{ccc}-21&1.5\\12&-41\\\end{array}\right]

B-CF=\left[\begin{array}{ccc}23&6.5\\-6&44\\\end{array}\right]

∴ the value of B - CF is \left[\begin{array}{ccc}23&6.5\\-6&44\\\end{array}\right]

<em>I hope this helps....</em>

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