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3 years ago

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PLZ HELP

1 answer:

Roman55 [17]3 years ago

4 0

She has 65 water bottles....and she is buying boxes of water bottles (x) that hold 16 water bottles each. She can only buy up to 5 water bottles.

f(x) = 16x + 65...where f(x) = total water bottles and x = number of boxes
x < = 5.....here is ur restriction

the range is ur f(x) values which represent the total number of water bottles...keep in mind, she can only buy 5. Therefore, the numbers that can be subbing in for x are 0 thru 5.

So when subbing in 0 thru 5 for x, the results for f(x) are :
{ 65,81,97,113,129 } <== ur range

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Answer:

a) A response of 8.9 represents the 92nd percentile.

b) A response of 6.6 represents the 62nd percentile.

c) A response of 4.4 represents the first quartile.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 5.9

Standard Deviation, σ = 2.2

We assume that the distribution of response is a bell shaped distribution that is a normal distribution.

Formula:

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

a) We have to find the value of x such that the probability is 0.92

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$P( X < x) = P( z < \displaystyle\frac{x - 5.9}{2.2})=0.92$

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

$P(z$

$\displaystyle\frac{x - 5.9}{2.2} = 1.405\\x = 8.991 \approx 8.9$

A response of 8.9 represents the 92nd percentile.

b) We have to find the value of x such that the probability is 0.62

P(X < x)

$P( X < x) = P( z < \displaystyle\frac{x - 5.9}{2.2})=0.62$

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

$P(z$

$\displaystyle\frac{x - 5.9}{2.2} = 0.305\\x = 6.571 \approx 6.6$

A response of 6.6 represents the 62nd percentile.

c) We have to find the value of x such that the probability is 0.25

P(X < x)

$P( X < x) = P( z < \displaystyle\frac{x - 5.9}{2.2})=0.25$

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

$P(z$

$\displaystyle\frac{x - 5.9}{2.2} = -0.674\\x = 4.4172 \approx 4.4$

A response of 4.4 represents the first quartile.

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