Mathematics, 16.03.2020 18:30, GhostElite6383

The annual per capita consumption of bottled water was 34.5 gallons. Assume that the per capita consumption of bottled water is approximately normally distributed with a mean of 34.5 and a standard deviation of 11 gallons. a. What is the probability that someone consumed more than 35 gallons of bottled water? b. What is the probability that someone consumed between 25 and 35 gallons of bottled water? c. What is the probability that someone consumed less than 25 gallons of bottled water? d. 97.5% of people consumed less than how many gallons of bottled water?

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answered: jahyra11

a) 0.4819

b) 0.3242

c) 0.1939

d) 56.06 gallons

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 34.5 gallons

Standard Deviation, σ = 11 gallons

We are given that the distribution of consumption of water is a bell shaped distribution that is a normal distribution.

Formula:

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

a) P(consumed more than 35 gallons of bottled water)

P(x > 35)

$P( x 35) = P( z \displaystyle\frac{35 - 34.5}{11}) = P(z 0.04545)$

$= 1 - P(z \leq 0.04545)$

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

$P(x 35) = 1 - 0.5181 = 0.4819 = 48.19\%$

b) P(consumed between 25 and 35 gallons of bottled water)

$P(25 \leq x \leq 35) \\\\= P(\displaystyle\frac{25 - 34.5}{11} \leq z \leq \displaystyle\frac{35-34.5}{11}) \\\\= P(-0.8636 \leq z \leq 0.0454)\\\\= P(z \leq 0.0454) - P(z < -0.8636)\\= 0.5181- 0.1939 =0.3242= 32.42\%$

$P(25 \leq x \leq 35) = 32.42\%$

c) P(consumed less than 25 gallons of bottled water)

$P( x < 25) = P( z < \displaystyle\frac{25 - 34.5}{11}) = P(z < -0.8636)$

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

$P(x < 25) = 0.1939= 19.39\%$

d) We have to find the value of x such that the probability is 0.975

$P( X < x) = P( z < \displaystyle\frac{x - 34.5}{1})=0.975$

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

$\displaystyle\frac{x - 34.5}{11} = 1.960\\\\x = 56.06$

97.5% of people consumed less than 56.06 gallons of bottled water.

answered: jescanarias22

(a) 0.48006

(b) 0.3251

(d) 97.5% of people consumed less than 56 gallons of bottled water.

Step-by-step explanation:

We are given that the per capita consumption of bottled water is approximately normally distributed with a mean of 34.5 and a standard deviation of 11 gallons.

Let X = per capita consumption of bottled water

So, X ~ N( $\mu = 34.5,\sigma^{2} =11^{2}$ )

The z score probability distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

(a) Probability that someone consumed more than 35 gallons of bottled water = P(X > 35)

P(X > 35) = P( $\frac{X-\mu}{\sigma}$ > $\frac{35-34.5}{11}$ ) = P(Z > 0.05) = 1 - P(Z $\leq$ 0.05)

= 1 - 0.51994 = 0.48006

(b) Probability that someone consumed between 25 and 35 gallons of bottled water = P(25 < X < 35)

P(25 < X < 35) = P(X < 35) - P(X $\leq$ 25)

P(X < 35) = P( $\frac{X-\mu}{\sigma}$ < $\frac{35-34.5}{11}$ ) = P(Z < 0.05) = 0.51994

P(X $\leq$ 25) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{25-34.5}{11}$ ) = P(Z $\leq$ -0.86) = 1 - P(Z < 0.86)

= 1 - 0.80511 = 0.19489

Therefore, P(25 < X < 35) = 0.51994 - 0.19489 = 0.3251

P(X < 25) = P( $\frac{X-\mu}{\sigma}$ < $\frac{25-34.5}{11}$ ) = P(Z < -0.86) = 1 - P(Z $\leq$ 0.86)

= 1 - 0.80511 = 0.19489

(d) We have to find that 97.5% of people consumed less than how many gallons of bottled water, which means ;

P(X < x) = 0.975

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-34.5}{11}$ ) = 0.975

P(Z > $\frac{x-34.5}{11}$ ) = 0.975

Now in the z table the critical value of X which have an area less than 0.975 is 1.96, i.e.;

$\frac{x-34.5}{11}$ = 1.96

- 34.5 = $1.96 \times 11$

= 34.5 + 21.56 = 56.06 ≈ 56 gallons of bottled water

So, 97.5% of people consumed less than 56 gallons of bottled water.

answered: Guest

the reporter's right

step-by-step explanation:

1/4 = .25 / 25%

answered: Guest

ummm seems that u didnt post the attachement !

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The annual per capita consumption of bottled water was 34.5 gallons. Assume that the per capita cons...