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I am Lyosha [343]
3 years ago
7

A researcher takes a simple random sample of 83 homes in a city and finds that the mean square footage is 1628 with a standard d

eviation of 243 square feet. Which interval is the 90% confidence interval for the mean square footage of all the homes in that city?
ANSWER:



(1584 ft2, 1672 ft2)
Mathematics
1 answer:
8090 [49]3 years ago
3 0

Answer:

Hello

Step-by-step explanation:

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8 0
3 years ago
The pregnancy length in days for a population of new mothers can be approximated by a normal distribution with a mean of days an
solniwko [45]

Answer:

(a) 283 days

(b) 248 days

Step-by-step explanation:

The complete question is:

The pregnancy length in days for a population of new mothers can be approximated by a normal distribution with a mean of 268 days and a standard deviation of 12 days. ​(a) What is the minimum pregnancy length that can be in the top 11​% of pregnancy​ lengths? ​(b) What is the maximum pregnancy length that can be in the bottom ​5% of pregnancy​ lengths?

Solution:

The random variable <em>X</em> can be defined as the pregnancy length in days.

Then, from the provided information X\sim N(\mu=268, \sigma^{2}=12^{2}).

(a)

The minimum pregnancy length that can be in the top 11​% of pregnancy​ lengths implies that:

P (X > x) = 0.11

⇒ P (Z > z) = 0.11

⇒ <em>z</em> = 1.23

Compute the value of <em>x</em> as follows:

z=\frac{x-\mu}{\sigma}\\\\1.23=\frac{x-268}{12}\\\\x=268+(12\times 1.23)\\\\x=282.76\\\\x\approx 283

Thus, the minimum pregnancy length that can be in the top 11​% of pregnancy​ lengths is 283 days.

(b)

The maximum pregnancy length that can be in the bottom ​5% of pregnancy​ lengths implies that:

P (X < x) = 0.05

⇒ P (Z < z) = 0.05

⇒ <em>z</em> = -1.645

Compute the value of <em>x</em> as follows:

z=\frac{x-\mu}{\sigma}\\\\-1.645=\frac{x-268}{12}\\\\x=268-(12\times 1.645)\\\\x=248.26\\\\x\approx 248

Thus, the maximum pregnancy length that can be in the bottom ​5% of pregnancy​ lengths is 248 days.

8 0
3 years ago
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