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svetoff [14.1K]
2 years ago
5

The pucks used by the National Hockey League for ice hockey must weigh between and ounces. Suppose the weights of pucks produced

at a factory are normally distributed with a mean of ounces and a standard deviation of ounces. What percentage of the pucks produced at this factory cannot be used by the National Hockey League? Round your answer to two decimal places.
Mathematics
1 answer:
Dahasolnce [82]2 years ago
4 0

Answer:

P(5.5

And we can find this probability using the normal standard distribution or excel and we got:

P(-2.769

Step-by-step explanation:

For this case we assume the following complete question: "The pucks used by the National Hockey League for ice hockey must weigh between 5.5 and 6 ounces. Suppose the weights of pucks produced at a factory are normally distributed with a mean of 5.86 ounces and a standard deviation of 0.13ounces. What percentage of the pucks produced at this factory cannot be used by the National Hockey League? Round your answer to two decimal places. "

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 weights of a population, and for this case we know the distribution for X is given by:

X \sim N(5.86,0.13)  

Where \mu=5.86 and \sigma=0.13

We are interested on this probability

P(5.5

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(5.5

And we can find this probability using the normal standard distribution or excel and we got:

P(-2.769

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Fudgin [204]

Answer:

A. $301

B. $721

Step-by-step explanation:

Let $x be the amount of money they raised.

Rowena tried to put the $1 bills into two equal piles and found one left over at the end, then

x=2q_1+1

Polly tried to put the $1 bills into three equal piles and found one left over at the end, then

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Frustrated, they tried 4, 5, and 6 equal piles and each time had $1 left over, then

x=4q_3+1\\ \\x=5q_4+1\\ \\x=6q_5+1

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Now, the number x=60n+1 should be divisible by 7 and must be greater than 500.

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x=60\cdot 11+1=661 is not divisible by 7.

When n = 12,

x=60\cdot 12+1=721 is divisible by 7.

B. The least amount of money they could have raised is $721

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