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(d) A drinks machine dispenses coffee into cups. A sign on the machine indicates that each cup contains 100ml of coffee. The mac

Marysya12 [62]

Answer:

The standard deviation of the amount of coffee dispensed per cup in ml is 3.91.

Step-by-step explanation:

Let the random variable X denote the amount of coffee dispensed by the machine.

It is provided that the random variable, X is normally distributed with mean, μ = 105 ml/cup and standard deviation, σ.

It is also provided that a sign on the machine indicates that each cup contains 100 ml of coffee.

And 10% of the cups contain less than the amount stated on the sign.

To compute the probabilities of a normally distributed random variable, first convert the raw score to a z-score,

$z=\frac{X-\mu}{\sigma}$

This implies that:

P (X < 100) = 0.10

⇒ P (Z < z) = 0.10

The value of z for the above probability is, z = -1.28.

*Use a z-table

Compute the value of standard deviation as follows:

$z=\frac{X-\mu}{\sigma}$

$-1.28=\frac{100-105}{\sigma}$

$\sigma=\frac{-5}{-1.28}$

$=3.90625\\\\\approx 3.91$

Thus, the standard deviation of the amount of coffee dispensed per cup in ml is 3.91.

8 0

3 years ago

The sum of four consecutive numbers is 626. Find the four numbers

iogann1982 [59]

155,156,157,158 divide 626 by 4 and you’ll get the middle number which is 156.5

3 0

3 years ago

Read 2 more answers

PLEASE HELP the information is below

liq [111]

Answer:

yellow: x=6

red: y=4

blue: y=-2

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

An equivalent equation for (a^1/2 times b^1/3)^1/2

zzz [600]

The product can be rewritten as
.. a^(1/4)*b^(1/6)

3 0

3 years ago

An article in Fire Technology describes the investigation of two different foam expanding agents that can be used in the nozzles

kvv77 [185]

Answer:

a) There is no evidence to support the claim that there is no difference in mean foam expansion of these two agents.

b) P=0

c) 90% CI

$-3.2406\leq \mu_1-\mu_2 \leq -2.2614$

This CI tells us that there is a 90% confidence that the real value of the difference between the means is between this two values. We see that both are negative values, so the value 0 is left out of the interval.

That means we can be almost sure both means don't have the same value, confirming the results of the previous test.

Step-by-step explanation:

The null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a: \mu_1-\mu_2\neq 0$

The significance level is 0.10 and it will be a two-tailed test.

The difference of the sample means is:

$M_d=M_1-M_2=4.340-7.091=-2.751$

As the sample size is equal for both samples, the estimated standard error of the difference between means is calculated as:

$s_M_d=\sqrt{\frac{s_1^2+s_2^2}{N}}= \sqrt{\frac{0.508^2+0.430^2}{5}}=\sqrt{\frac{0.442964}{5} }= \sqrt{0.0886}=0.2976$

Then, the statistic z is:

$z=\frac{M_d-(\mu_1-\mu_2)}{s_M_d}=\frac{-2.751-0}{0.2976}=-9.24\\\\P(|z|>9.24)=0$

The P-value (P=0) is much lower than the significance level, so the null hypothesis is rejected. The means are different.

For a 90% confidence interval for the difference of the means, we use a z=1.645.

Then the confidence interval is defined as:

$M_d-z*s_M_d\leq \mu_1-\mu_2 \leq M_d-z*s_M_d\\\\-2.751-1.645*0.2976\leq \mu_1-\mu_2 \leq -2.751+1.645*0.2976\\\\-3.2406\leq \mu_1-\mu_2 \leq -2.2614$

This CI tells us that there is a 90% confidence that the real value of the difference between the means is between this two values. We see that both are negative values, so the value 0 is left out of the interval.

That means we can be almost sure both means don't have the same value.

5 0

3 years ago