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Marysya12 [62]
2 years ago
8

Laboratory tests show that the lives of

Mathematics
1 answer:
SIZIF [17.4K]2 years ago
6 0
<h3>Answer:  34%</h3>

This result is approximate.

=========================================================

Explanation:

mu = 750 = mean

sigma = 75 = standard deviation

The raw scores or x values are x = 750 and x = 825

Let's compute the z score for each x value

z = (x - mu)/sigma

z = (750 - 750)/75

z = 0

and

z = (x - mu)/sigma

z = (825 - 750)/75

z = 1

Therefore P(750 ≤ x ≤ 825) is equivalent to P(0 ≤ z ≤ 1) in this context.

Use a z score table to determine that

P(z ≤ 0) = 0.5

P(z ≤ 1) = 0.84314 approximately

So,

P(a ≤ z ≤ b) = P(z ≤ b) - P(z ≤ a)

P(0 ≤ z ≤ 1) = P(z ≤ 1) - P(z ≤ 0)

P(0 ≤ z ≤ 1) = 0.84314 - 0.5

P(0 ≤ z ≤ 1) = 0.34314 approximately

The value 0.34314 then converts to 34.314% which rounds to <u>34%</u>

Or you could use the empirical rule as shown below. The pink section on the right is marked <u>34%</u> which is approximate. This pink section is between z = 0 and z = 1.

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The mean cost of domestic airfares in the United States rose to an all-time high of $385 per ticket (Bureau of Transportation St
saveliy_v [14]

Answer:

a)0.067

b)0.111

c)0.612

d)$687.28

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = $385

Standard Deviation, σ = $110

We are given that the distribution of domestic airfares is a bell shaped distribution that is a normal distribution.

Formula:

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

a) P(domestic airfare is $550 or more)

P(x > 550)

P( x > 550) = P( z > \displaystyle\frac{550 - 385}{110}) = P(z > 1.5)

= 1 - P(z \leq 1.5)

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

P(x > 550) = 1 - 0.933 = 0.067= 6.7\%

b) P(domestic airfare is $250 or less)

P(x \leq 250) = P(z \leq \displaystyle\frac{250-385}{110}) = P(z \leq -1.22)

Calculating the value from the standard normal table we have,

P( x \leq 250) = 0.111 = 11.1\%

c))P(domestic airfare is between $300 and $500)

P(300 \leq x \leq 500) = P(\displaystyle\frac{300 - 385}{110} \leq z \leq \displaystyle\frac{500-385}{110}) = P(-0.77 \leq z \leq 1.04)\\\\= P(z \leq 1.04) - P(z < -0.77)\\= 0.851 - 0.239 = 0.612 = 61.2\%

P(300 \leq x \leq 500) = 61.2\%

d) P(X=x) = 0.03

We have to find the value of x such that the probability is 0.03.

P(X > x)

P( X > x) = P( z > \displaystyle\frac{x - 385}{110})=0.03

= 1 -P( z \leq \displaystyle\frac{x - 385}{110})=0.03

=P( z \leq \displaystyle\frac{x - 385}{110})=0.997

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

\displaystyle\frac{x - 385}{110} = 2.748\\x = 687.28

Hence, the domestic fares must be $687.28 or greater for them to lie in the highest 3%.

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Select an expression equal to 6(4q + 8r)​
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