Ask question

Ask question

All categories

3 years ago

7

A poultry farmer wishes to estimate the average incubation period (the number of days between a hen laying her egg and the time

the egg hatches) for eggs on his farm. He plans to take a sample and make a 98% confidence interval, and would like a margin of error of half a day. It is known that the distribution of incubation lengths has a standard deviation of 1.5 days. How many eggs does he need to sample to create the desired interval

1 answer:

sergiy2304 [10]3 years ago

8 0

Answer:

49 eggs.

Step-by-step explanation:

The formula for Margin of Error =

z × standard deviation/√number of samples

z = z score of 98% confidence interval = 2.326

Margin of Error = Half a day = 1/2day = 0.5 day

Standard deviation = 1.5 days

Number of samples = number of eggs he needs to sample = unknown.

Imputing these above values into the formula

Margin of Error = z × standard deviation/√number of samples

0.5 = 2.326 × 1.5/√n

Cross Multiply

0.5 × √n = 2.326 × 1.5

√n = 2.326 × 1.5/0.5

√n = 3.489/0.5

√n = 6.978

Square both sides

(√n)² = 6.978²

n = 48.692484

n ≈ Approximately to the nearest whole number = 49

Therefore, the number of eggs he needs to sample to create the desired interval is approximately to the nearest whole number 49 eggs

You might be interested in

What is the slope please help

mylen [45]

Slope: rise over run
4 up and 2 to left
4/2 = 2, the slope is 2

8 0

3 years ago

The SAT scores have an average of 1200 with a standard deviation of 60. A sample of 36 scores is selected. a) What is the probab

LiRa [457]

Answer:

a) 0.0082

b) 0.9987

c) 0.9192

d) 0.5000

e) 1

Step-by-step explanation:

The question is concerned with the mean of a sample.

From the central limit theorem we have the formula:

$z=\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$

a) $z=\frac{1224-1200}{\frac{60}{\sqrt{36} } }=2.40$

The area to the left of z=2.40 is 0.9918

The area to the right of z=2.40 is 1-0.9918=0.0082

$\therefore P(\bar X\:>\:1224)=0.0082$

b) $z=\frac{1230-1200}{\frac{60}{\sqrt{36} } }=3.00$

The area to the left of z=3.00 is 0.9987

$\therefore P(\bar X\:$

c) The z-value of 1200 is 0

The area to the left of 0 is 0.5

$z=\frac{1214-1200}{\frac{60}{\sqrt{36} } }=1.40$

The area to the left of z=1.40 is 0.9192

The probability that the sample mean is between 1200 and 1214 is

$P(1200\:$

d) From c) the probability that the sample mean will be greater than 1200 is 1-0.5000=0.5000

e) $z=\frac{73.46-1200}{\frac{60}{\sqrt{36} } }=-112.65$

The area to the left of z=-112.65 is 0.

The area to the right of z=-112.65 is 1-0=1

5 0

3 years ago

The original cost of a pair of shoes is 850dhs. The store offers a markup of 30%. Find the retail price of pair of shoes.

mamaluj [8]

Answer:

Step-by-step explanation:

7 0

2 years ago

On Monday your run on a treadmill for 1/2 hour at x miles. how do you solve it

fgiga [73]

You ran 6 miles on the treadmill on Monday.

7 0

2 years ago

Sasha hikes into a canyon. She takes a break at the rest stations every 23 of a mile she descends.

katrin [286]

Answer:

hi bye

Step-by-step explanation:

5 0

3 years ago