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2 years ago

A survey of students' favorite sports was taken from a random sample of

Mathematics

1 answer:

White raven [17]2 years ago

6 0

Answer:

129

Step-by-step explanation:

Considering the survey to be representative, you can simply multiply the share of students p preferring “Track & Field” with the whole school population at the same time to estimate the number of such students in the whole school.

First we need to find the relative share p of such answers in the study by dividing it by the sum of answers, assuming that the table is complete for that random sample:

p = 4/(8 + 5 + 4) = 4/17

Then for the whole school we get 550 p ≈ 129.4

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3 years ago

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2 years ago

The attendance at baseball games at a certain stadium is normally distributed, with a mean of 30,000 and a standard deviation of

lakkis [162]

Answer:

a) 0.964

b) 0.500

c) 0.885

d) $x \geq 32467.5$

e) 0.997

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 30000

Standard Deviation, σ = 1500

We are given that the distribution of attendance at stadium is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(attendance is greater than 27,300)

P(x > 27300)

$P( x > 27300) = P( z > \displaystyle\frac{27300 - 30000}{1500}) = P(z > -1.8)$

$= 1 - P(z \leq -1.8)$

Calculation the value from standard normal z table, we have,

$P(x > 27300) = 1 - 0.036 = 0.964 = 96.4\%$

b) P(attendance greater than or equal to 30000)

$P(x > 30000) = P(z > \displaystyle\frac{30000-30000}{1500}) = P(z \geq 0)\\\\P( z \geq 0) = 1 - P(z \leq 0)$

Calculating the value from the standard normal table we have,

$1 - 0.500 = 0.500 = 50\%$

c) P(attendance between 27000 and 32000)

$P(27000 \leq x \leq 32000) = P(\displaystyle\frac{27000 - 30000}{1500} \leq z \leq \displaystyle\frac{32000-30000}{1500}) = P(-2 \leq z \leq 1.33)\\\\= P(z \leq 1.33) - P(z < -2)\\= 0.908 - 0.023 = 0.885 = 88.5\%$