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pychu [463]
3 years ago
11

Trucks in a delivery fleet travel a mean of 100 miles per day with a standard deviation of 37 miles per day. The mileage per day

is distributed normally. Find the probability that a truck drives between 166 and 177 miles in a day. Round your answer to four decimal places.
Mathematics
1 answer:
timama [110]3 years ago
5 0

Answer: the probability that a truck drives between 166 and 177 miles in a day is 0.0187

Step-by-step explanation:

Since mileage of trucks per day is distributed normally, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = mileage of truck

µ = mean mileage

σ = standard deviation

From the information given,

µ = 100 miles per day

σ = 37 miles miles per day

The probability that a truck drives between 166 and 177 miles in a day is expressed as

P(166 ≤ x ≤ 177)

For x = 166

z = (166 - 100)/37 = 1.78

Looking at the normal distribution table, the probability corresponding to the z score is 0.9625

For x = 177

z = (177 - 100)/37 = 2.08

Looking at the normal distribution table, the probability corresponding to the z score is 0.9812

Therefore,

P(166 ≤ x ≤ 177) = 0.9812 - 0.9625 = 0.0187

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Step-by-step explanation:

Let <em>X</em> denote the heights of women in the USA.

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(a)

Compute the probability that the sample mean is greater than 63 inches as follows:

P(\bar X>63)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{63-64}{3/\sqrt{6}})\\\\=P(Z>-0.82)\\\\=P(Z

Thus, the probability that the sample mean is greater than 63 inches is 0.2061.

(b)

Compute the probability that a randomly selected woman is taller than 66 inches as follows:

P(X>66)=P(\frac{X-\mu}{\sigma}>\frac{66-64}{3})\\\\=P(Z>0.67)\\\\=1-P(Z

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(c)

Compute the probability that the mean height of a random sample of 100 women is greater than 66 inches as follows:

P(\bar X>66)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{66-64}{3/\sqrt{100}})\\\\=P(Z>6.67)\\\\\ =0

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