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scoundrel [369]
2 years ago
5

The length of time a person takes to decide which shoes to purchase is normally distributed with a mean of 8.54 minutes and a st

andard deviation of 1.91. Find the probability that a randomly selected individual will take less than 6 minutes to select a shoe purchase. Is this outcome unusual?
Mathematics
1 answer:
AnnZ [28]2 years ago
4 0

Answer:  0.0918, it is not unusual.

Step-by-step explanation:

Given : The length of time a person takes to decide which shoes to purchase is normally distributed with a mean of 8.54 minutes and a standard deviation of 1.91.

i.e. \mu=8.54 minutes and \sigma= 1.91 minutes

Let x denotes the length of time a person takes to decide which shoes to purchase.

Formula : z=\dfrac{x-\mu}{\sigma}

Then, the probability that a randomly selected individual will take less than 6 minutes to select a shoe purchase will be :-

\text{P-value=}P(x

Thus , the required probability = 0.0918

Since, P-value (0.0918) >0.05 , it means this outcome is not unusual.

[Note : When a outcome is unusual then the probability of its happening is less than or equal to 0.05. ]

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Answer:

m+7

Step-by-step explanation:

If she ran 7 minutes longer and m represents the amount of time she ran, then it would just be addition.

Let's say m=60 minutes. She would've ran for 67 minutes, which is also m+7.

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Tim has an after-school delivery service that he provides for several small retailers in town. He uses his bicycle and charges $
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Answer:

<em>$4.84</em>

<em></em>

Step-by-step explanation:

Given that

For delivery within 1\frac{1}2 mi, charges = $1.25

For delivery within 1\frac{1}2 mi - 1\frac{3}4 mi, charges = $1.70

For delivery within 1\frac{3}4 mi - 2 mi, charges = $2.15

and

so on.

i.e.

For delivery within 2 mi - 2\frac{1}4 mi, charges = $2.60

For delivery within 2\frac{1}4 mi - 2\frac{1}2 mi, charges = $3.05

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So, every 0.25 mi or \frac{1}4 mi increase in distance, there is an increase of $0.45 in the charges.

It is given that there is increase of 10% in the rates.

i.e. for every 0.25 mi increase in distance, there is an increase of 0.45*1.10 = $0.495 in the charges.

The distance of 3\frac{1}8 miles lies within the range 3 mi - 3\frac{1}4 mi.

So actual charges after increase of 10%

\Rightarrow \$4.40 \times \dfrac{110}{100}\\\Rightarrow \$4.40 \times 1.10\\\Rightarrow \bold{\$4.84}

6 0
2 years ago
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