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kirill115 [55]
2 years ago
10

Given that a 90% confidence interval for the mean height of all adult males in Idaho measured in inches was [62.532, 76.478]. Wh

at was the point estimate used to estimate the mean height of all adult males in Idaho?
Mathematics
1 answer:
NNADVOKAT [17]2 years ago
3 0

Answer:

The point estimate used to estimate the mean height of all adult males in Idaho is 69.505 inches.

Step-by-step explanation:

Each confidence interval has two bounds, the lower bound and the upper bound. The points estimate used to estimate the mean is the halfway point between those two bounds, that is, the sum of those two bounds divided by two.

In this problem, we have that:

Lower bound: 62.532

Upper bound: 76.478

Point estimate: (62.532 + 76.478)/2 = 69.505

The point estimate used to estimate the mean height of all adult males in Idaho is 69.505 inches.

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Choose Yes or No to tell whether the expressions are equivalent. 1. 4(5c + 3) and 9c + 7 yes or no 2. 10f – 10 and 2(8f – 5) yes
Juli2301 [7.4K]

Answer:

1) No

2) No

3)Yes

4)No

Step-by-step explanation:

Distributive property: a(b + c) =a*b + a*c

1) 4(5c + 3) = (4*5c) + 4*3

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2) 2(8f - 5) = (2*8f) - (2*5)

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5 0
3 years ago
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Zarrin [17]

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Step-by-step explanation:

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2 years ago
Math scores on the SAT exam are normally distributed with a mean of 514 and a standard deviation of 118. If a recent test-taker
LuckyWell [14K]

Answer:

Probability that the student scored between 455 and 573 on the exam is 0.38292.

Step-by-step explanation:

We are given that Math scores on the SAT exam are normally distributed with a mean of 514 and a standard deviation of 118.

<u><em>Let X = Math scores on the SAT exam</em></u>

So, X ~ Normal(\mu=514,\sigma^{2} =118^{2})

The z score probability distribution for normal distribution is given by;

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

where, \mu = population mean score = 514

           \sigma = standard deviation = 118

Now, the probability that the student scored between 455 and 573 on the exam is given by = P(455 < X < 573)

       P(455 < X < 573) = P(X < 573) - P(X \leq 455)

       P(X < 573) = P( \frac{X-\mu}{\sigma} < \frac{573-514}{118} ) = P(Z < 0.50) = 0.69146

       P(X \leq 2.9) = P( \frac{X-\mu}{\sigma} \leq \frac{455-514}{118} ) = P(Z \leq -0.50) = 1 - P(Z < 0.50)

                                                         = 1 - 0.69146 = 0.30854

<em>The above probability is calculated by looking at the value of x = 0.50 in the z table which has an area of 0.69146.</em>

Therefore, P(455 < X < 573) = 0.69146 - 0.30854 = <u>0.38292</u>

Hence, probability that the student scored between 455 and 573 on the exam is 0.38292.

7 0
3 years ago
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