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mixas84 [53]
4 years ago
11

If male and female college students work an average of 15 hours per week during the school year and the standard deviation is 5.

4 for male students and 2.3 for female students, what may we conclude?
Mathematics
2 answers:
charle [14.2K]4 years ago
5 0

Answer: the male sample is more spread than the female sample

Step-by-step explanation:

The mean is the same for both genres, but the standard deviation form males is bigger than the one for females.

This means that the data of the male students is more spread than the female data (so there are more male students that study more hours, but there also are more male students that study less hours).

So the general conclusion is that the male sample is more spread than the female sample

ki77a [65]4 years ago
3 0
The variance of hours males spend working is almost two times as large as females. 60% of those females are more likely to work between 12.7 to 17.3 hours while 60% of the males work between 9.6 to 20.3 hours based on one standard deviation.
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Let the Dulcina's collection be 'x'

Let the Tremaine collection be 'x-39'

x + x - 39 =129

2x = 129 +39

2x = 168

x = 168/2

x = 84

Dulcina's collection = x = 84

Tremaine's collection = x - 39 = 84 - 39 = 45

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Which list of numbers is shown on the number line below?
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2 years ago
A trapezoid has an area of 780 square meters. The height of the trapezoid is 40 ​meters, and the length of the longer base is tw
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3 years ago
i need help on this problem i know how to do this types of problem but i dont get this problem i need hep thank you
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B: 72

Step-by-step explanation:

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8 0
3 years ago
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The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per
ICE Princess25 [194]

Answer:

1.76% probability that in one hour more than 5 clients arrive

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given time interval.

The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per hour.

This means that \mu = 2

What is the probability that in one hour more than 5 clients arrive

Either 5 or less clients arrive, or more than 5 do. The sum of the probabilities of these events is decimal 1. So

P(X \leq 5) + P(X > 5) = 1

We want P(X > 5). So

P(X > 5) = 1 - P(X \leq 5)

In which

P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353

P(X = 1) = \frac{e^{-2}*2^{1}}{(1)!} = 0.2707

P(X = 2) = \frac{e^{-2}*2^{2}}{(2)!} = 0.2707

P(X = 3) = \frac{e^{-2}*2^{3}}{(3)!} = 0.1804

P(X = 4) = \frac{e^{-2}*2^{4}}{(4)!} = 0.0902

P(X = 5) = \frac{e^{-2}*2^{5}}{(5)!} = 0.0361

P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1353 + 0.2702 + 0.2702 + 0.1804 + 0.0902 + 0.0361 = 0.9824

P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9824 = 0.0176

1.76% probability that in one hour more than 5 clients arrive

8 0
3 years ago
Read 2 more answers
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