Help please . What does ad equal

An article estimates, with 90% confidence, that the proportion of elementary school teachers who are female is 82%. The article

GREYUIT [131]

Answer: (80% , 85%)

Step-by-step explanation:

We know that, confidence interval for population proportion is given by :-

(p-E , p+E)

, where p = sample proportion , E = Margin of error.

Given : Proportion of elementary school teachers who are female = 82%.

The article also states the maximum error of their estimate = 3%.

Then, the 90% confidence interval for the proportion of elementary school teachers who are female will be :

(82%-2% , 82%+3%)

= (80% , 85%)

Hence, the resulting 90% confidence interval for the proportion of elementary school teachers who are female = (80% , 85%)

5 0

3 years ago

Wilma works at a bird sanctuary and stores birdseed in plastic containers. She has 2 small containers that hold 8 pounds of bird

Amanda [17]

Answer:

60 lbs left over

Step-by-step explanation: Since the small containers hold 8 lbs each you multiply 8 by 2 since there are 2 containers. Same for the large, multiply 9 by 12 because each container holds 12 lbs and there is 9 containers. Then, you subtract 4 from that total because she already used 4 lbs.

x=(2*8)+(9*12)-4 Multiply

x=16+108-4 Simplify

x=120

So, before she fills the bird feeders she has 120 lbs of bird seed. There are 30 bird feeders and since they hold 2 lbs each you multiply 30 and 2. Then you subtract that from the 120 lbs because she is filling the bird feeders with that seed.

120- (30*2)

120-60= 60 lbs left over

3 0

3 years ago

An automobile manufacturer claims that its jeep has a 53.9 miles/gallon (MPG) rating. An independent testing firm has been contr

Advocard [28]

Answer:

Value of the test statistic, $z_{test} = - 1.17$

Step-by-step explanation:

Null hypothesis, $H_{0}: \mu = 53.9$

Alternative hypothesis, $H_{a} : \mu \neq 53.9$

Sample mean, $\bar{X} = 53.7$

Sample size, n = 110

Standard deviation, $\sigma = 1.8$

Significance level, $\alpha = 0.02$

The value of the test statistics is given by the formula:

$z_{test} = \frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt{n} } } \\z_{test} = \frac{53.7 - 53.9}{\frac{1.8}{\sqrt{110} } } \\z_{test} = - 1.17$

7 0

2 years ago

The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm197.5 cm and a standard deviation

fiasKO [112]

Answer:

a) 5.37% probability that an individual distance is greater than 210.9 cm

b) 75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c) Because the underlying distribution is normal. We only have to verify the sample size if the underlying population is not normal.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

$\mu = 197.5, \sigma = 8.3$

a. Find the probability that an individual distance is greater than 210.9 cm

This is 1 subtracted by the pvalue of Z when X = 210.9. So

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

$Z = \frac{210.9 - 197.5}{8.3}$

$Z = 1.61$

$Z = 1.61$ has a pvalue of 0.9463.

1 - 0.9463 = 0.0537

5.37% probability that an individual distance is greater than 210.9 cm.

b. Find the probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

Now $n = 15, s = \frac{8.3}{\sqrt{15}} = 2.14$

This probability is 1 subtracted by the pvalue of Z when X = 196. Then

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

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{196 - 197.5}{2.14}$

$Z = -0.7$

$Z = -0.7$ has a pvalue of 0.2420.

1 - 0.2420 = 0.7580

75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

The underlying distribution(overhead reach distances of adult females) is normal, which means that the sample size requirement(being at least 30) does not apply.

5 0

3 years ago

Double a number decreased by 4

Wittaler [7]

Answer:

2x-5 if u want u can substitute any other variables and constant

3 0

2 years ago

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