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Ber [7]
3 years ago
13

The owner of a fish market has an assistant who has determined that the weights of catfish are normally distributed, with mean o

f 3.2 pounds and standard deviation of 0.8 pound. What percentage of samples of 4 fish will have sample means between 3.0 and 4.0 pounds
Mathematics
1 answer:
prohojiy [21]3 years ago
6 0

Answer:

The percentage of samples of 4 fish will have sample means between 3.0 and 4.0 pounds is 66.87%

Step-by-step explanation:

For a normal random variable with mean Mu = 3.2 and standard deviation sd = 0.8 there is a distribution of the sample mean (MX) for samples of size 4, given by:

Z = (MX - Mu) / sqrt (sd ^ 2 / n) = (MX - 3.2) / sqrt (0.64 / 4) = (MX - 3.2) / 0.4

For a sample mean of 3.0, Z = (3 - 3.2) / 0.4 = -0.5

For a sample mean of 3.0, Z = (4 - 3.2) / 0.4 = 2.0

P (3.2 <MX <4) = P (-0.5 < Z <2.0) = 0.6687.

The percentage of samples of 4 fish will have sample means between 3.0 and 4.0 pounds is 66.87%

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Answer:

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Step-by-step explanation:

Given:

KL ║ NM ,

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1. Consider triangle NLM. This is a right triangle, because NL ⊥ LM. In this triangle,

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Also

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2. Consider triangle NKL. This is a right triangle, because KN ⊥ NM . In this triangle,

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m\angle KNL=90^{\circ}-40^{\circ}=50^{\circ} (angles KNL and KLN are complementary).

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\sin \angle KNL=\dfrac{\text{opposite leg}}{\text{hypotenuse}}=\dfrac{KL}{LN}=\dfrac{KL}{45\tan 50^{\circ}}\\ \\KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08

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\cos \angle KNL=\dfrac{\text{adjacent leg}}{\text{hypotenuse}}=\dfrac{KN}{LN}=\dfrac{KN}{45\tan 50^{\circ}}\\ \\KN=45\tan 50^{\circ}\cos 50^{\circ}=45\sin 50^{\circ}\approx 34.47

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