1. In the figure at the right, it is given

A physical fitness association is including the mile run in its secondary- school fitness test. the time for this event for boys

liberstina [14]

Answer:

<u>The probability that a randomly selected boy in school can run the mile in less than 348 seconds is 1.1%.</u>

Step-by-step explanation:

1. Let's review the information provided to us to answer the question correctly:

μ of the time a group of boys run the mile in its secondary- school fitness test = 440 seconds

σ of the time a group of boys run the mile in its secondary- school fitness test = 40 seconds

2. Find the probability that a randomly selected boy in school can run the mile in less than 348 seconds.

Let's find out the z-score, this way:

z-score = (348 - 440)/40

z-score = -92/40 = -2.3

Now let's find out the probability of z-score = -2.3, using the table:

p (-2.3) = 0.0107

p (-2.3) = 0.0107 * 100

p (-2.3) = 1.1% (rounding to the next tenth)

<u>The probability that a randomly selected boy in school can run the mile in less than 348 seconds is 1.1%.</u>

4 0

2 years ago

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15. Suppose a box of 30 light bulbs contains 4 defective ones. If 5 bulbs are to be removed out of the box.

Naddika [18.5K]

Answer:

1) Probability that all five are good = 0.46

2) P(at most 2 defective) = 0.99

3) Pr(at least 1 defective) = 0.54

Step-by-step explanation:

The total number of bulbs = 30

Number of defective bulbs = 4

Number of good bulbs = 30 - 4 = 26

Number of ways of selecting 5 bulbs from 30 bulbs = $30C5 = \frac{30 !}{(30-5)!5!} \\$

$30C5 = 142506$ ways

Number of ways of selecting 5 good bulbs from 26 bulbs = $26C5 = \frac{26 !}{(26-5)!5!} \\$

$26C5 = 65780$ ways

Probability that all five are good = 65780/142506

Probability that all five are good = 0.46

2) probability that at most two bulbs are defective = Pr(no defective) + Pr(1 defective) + Pr(2 defective)

Pr(no defective) has been calculated above = 0.46

Pr(1 defective) = $\frac{26C4 * 4C1}{30C5}$

Pr(1 defective) = (14950*4)/142506

Pr(1 defective) =0.42

Pr(2 defective) = $\frac{26C3 * 4C2}{30C5}$

Pr(2 defective) = (2600 *6)/142506

Pr(2 defective) = 0.11

P(at most 2 defective) = 0.46 + 0.42 + 0.11

P(at most 2 defective) = 0.99

3) Probability that at least one bulb is defective = Pr(1 defective) + Pr(2 defective) + Pr(3 defective) + Pr(4 defective)

Pr(1 defective) =0.42

Pr(2 defective) = 0.11

Pr(3 defective) = $\frac{26C2 * 4C3}{30C5}$

Pr(3 defective) = 0.009

Pr(4 defective) = $\frac{26C1 * 4C4}{30C5}$

Pr(4 defective) = 0.00018

Pr(at least 1 defective) = 0.42 + 0.11 + 0.009 + 0.00018

Pr(at least 1 defective) = 0.54

3 0

3 years ago

Plzzzzzzzzzzzzzzzzzzzzzz neeed help dew nowwwwwwwwwwwwwwwwwwwwwwwwwww

vampirchik [111]

Answer:

64 teachers

Step-by-step explanation:

We can use ratios to solve

14 students 896 students

----------------- = -------------

1 teacher x teachers

Using cross products

14 * x = 896*1

Divide each side by 14

14x/14 = 896/14

x =64

4 0

3 years ago

The graph of the function f (x) is shown below. When f(x)=0,x=?

IgorLugansk [536]

Answer:

0

Step-by-step explanation:

7 0

3 years ago

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(1) An M&Ms factory produces 109,500,000 M&M’s every day. About ¼ of its M&Ms are blue. Find the number of blue M&am

Gemiola [76]

Answer:

27,375,000

Step-by-step explanation:

Simply divide the total by 4.

4 0

3 years ago