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Setler [38]
3 years ago
9

Neil is a drummer who purchases his drumsticks online. When practicing with the newest pair, he notices they feel heavier than u

sual. When he weighs one of the sticks, he finds that it is 2.33 oz. The manufacturer's website states that the average weight of each stick is 1.75 oz with a standard deviation of 0.22 oz. Assume that the weight of the drumsticks is normally distributed. What is the probability of the stick's weight being 2.33 oz or greater
Mathematics
1 answer:
Fynjy0 [20]3 years ago
5 0

Answer:

The probability of the stick's weight being 2.33 oz or greater is 0.0041 or 0.41%.

Step-by-step explanation:

Given:

Weight of a given sample (x) = 2.33 oz

Mean weight (μ) = 1.75 oz

Standard deviation (σ) = 0.22 oz

The distribution is normal distribution.

So, first, we will find the z-score of the distribution using the formula:

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

Plug in the values and solve for 'z'. This gives,

z=\frac{2.33-1.75}{0.22}=2.64

So, the z-score of the distribution is 2.64.

Now, we need the probability P(x\geq 2.33 )=P(z\geq  2.64).

From the normal distribution table for z-score equal to 2.64, the value of the probability is 0.9959. This is the area to the left of the curve or less than z-score value.

But, we need area more than the z-score value. So, the area is:

P(z\geq  2.64)=1-0.9959=0.0041=0.41\%

Therefore, the probability of the stick's weight being 2.33 oz or greater is 0.0041 or 0.41%.

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Step-by-step explanation:

<u>You need to find the difference:</u>

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The heights of men in a certain population follow a normal distribution with mean 69.7 inches and standard deviation 2.8 inches.
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Answer:

a) P(Y > 76) = 0.0122

b) i) P(both of them will be more than 76 inches tall) = 0.00015

   ii) P(Y > 76) = 0.0007

Step-by-step explanation:

Given - The heights of men in a certain population follow a normal distribution with mean 69.7 inches and standard deviation 2.8 inches.

To find - (a) If a man is chosen at random from the population, find

                    the probability that he will be more than 76 inches tall.

              (b) If two men are chosen at random from the population, find

                    the probability that

                    (i) both of them will be more than 76 inches tall;

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P(Y > 76) = P(Y - mean > 76 - mean)

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b)

(i)

P(both of them will be more than 76 inches tall) = (0.0122)²

                                                                           = 0.00015

⇒P(both of them will be more than 76 inches tall) = 0.00015

(ii)

Given that,

Mean = 69.7,

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P(Y > 76) = P(Y - mean > 76 - mean)

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                 = P(Z > 3.182)

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