In: Math

17. The lengths of a population of certain HULU shows I watch are normally distributed with a mean running time of 38 minutes and a standard deviation of 11.5.

- Find the percentage of shows with running times between 26 and 42 minutes.

2.Between what values would you expect to find the middle 80 %

3. Find the percentage of shows with running times below 47.5 minutes

4.Above what value would you expect to find the top 25 %?

5.Find the percentage of shows with running times above 18 minutes.

Part 1)

X ~ N ( µ = 38 , σ = 11.5 )

P ( 26 < X < 42 )

Standardizing the value

Z = ( X - µ ) / σ

Z = ( 26 - 38 ) / 11.5

Z = -1.0435

Z = ( 42 - 38 ) / 11.5

Z = 0.3478

P ( -1.04 < Z < 0.35 )

P ( 26 < X < 42 ) = P ( Z < 0.35 ) - P ( Z < -1.04
)

P ( 26 < X < 42 ) = 0.636 - 0.1484

**P ( 26 < X < 42 ) = 0.4876**

Part 2)

X ~ N ( µ = 38 , σ = 11.5 )

P ( a < X < b ) = 0.8

Dividing the area 0.8 in two parts we get 0.8/2 = 0.4

since 0.5 area in normal curve is above and below the mean

Area below the mean is a = 0.5 - 0.4

Area above the mean is b = 0.5 + 0.4

Looking for the probability 0.1 in standard normal table to
calculate critical value Z = -1.2816

Looking for the probability 0.9 in standard normal table to
calculate critical value Z = 1.2816

Z = ( X - µ ) / σ

-1.2816 = ( X - 38 ) / 11.5

a = 23.2616

1.2816 = ( X - 38 ) / 11.5

b = 52.7384

**P ( 23.2616 < X < 52.7384 ) = 0.8**

Part 3)

X ~ N ( µ = 38 , σ = 11.5 )

P ( X < 47.5 )

Standardizing the value

Z = ( X - µ ) / σ

Z = ( 47.5 - 38 ) / 11.5

Z = 0.8261

P ( ( X - µ ) / σ ) < ( 47.5 - 38 ) / 11.5 )

P ( X < 47.5 ) = P ( Z < 0.8261 )

P ( X < 47.5 ) = 0.7956

Percentage is 0.7956 * 100 = 79.56%

Part 4)

X ~ N ( µ = 38 , σ = 11.5 )

P ( X > x ) = 1 - P ( X < x ) = 1 - 0.25 = 0.75

To find the value of x

Looking for the probability 0.75 in standard normal table to
calculate critical value Z = 0.6745

Z = ( X - µ ) / σ

0.6745 = ( X - 38 ) / 11.5

**X = 45.7567**

P ( X > 45.7567 ) = 0.25

Part 5)

X ~ N ( µ = 38 , σ = 11.5 )

P ( X > 18 ) = 1 - P ( X < 18 )

Standardizing the value

Z = ( X - µ ) / σ

Z = ( 18 - 38 ) / 11.5

Z = -1.7391

P ( ( X - µ ) / σ ) > ( 18 - 38 ) / 11.5 )

P ( Z > -1.7391 )

P ( X > 18 ) = 1 - P ( Z < -1.7391 )

P ( X > 18 ) = 1 - 0.041

P ( X > 18 ) = 0.959

Percentage is 0.9590 * 100 = 95.90%

Suppose the lengths of the pregnancies of a certain animal are
approximately normally distributed with mean equals 136 days and
standard deviation equals 12 days.
What is the probability a random sample of size 19 will have a
mean gestation period within 8 days of the mean

Suppose the lengths of the pregnancies of a certain animal are
approximately normally distributed with mean
mu equals μ=197 daysand standard deviation sigma equals sσ=14
days. Complete parts (a) through (f) below.
(a) What is the probability that a randomly selected pregnancy
lasts less than 192 days?
The probability that a randomly selected pregnancy lasts less
than 192 days is approximately 0.3604 (Round to four decimal
places as needed.)
Interpret this probability. Select the correct choice below and
fill in...

Suppose the lengths of the pregnancies of a certain animal are
approximately normally distributed with mean mu equals 248 days and
standard deviation sigma equals 24 days. Complete parts (a)
through (f) below. (a) What is the probability that a randomly
selected pregnancy lasts less than 239 days? The probability that
a randomly selected pregnancy lasts less than 239 days is
approximately nothing. (Round to four decimal places as needed.)
Interpret this probability. Select the correct choice below and
fill...

Suppose the lengths of the pregnancies of a certain animal are
approximately normally distributed with mean= 169days and standard
deviation=14 days. complete parts (a) through (f) below.
(a) What is the probability that a randomly selected pregnancy
lasts less than 164 days?
The probability that a randomly selected pregnancy lasts less
than 164 days is approximately _. (Round to four decimal places as
needed.)
Interpret this probability. Select the correct choice below and
fill in the answer box within your...

Suppose the lengths of the pregnancies of a certain animal are
approximately normally distributed with mean μ=253 days and
standard deviation σ=17 days. Complete parts (a) through (f)
below.
(a) What is the probability that a randomly
selected pregnancy lasts less than 247 days?
The probability that a randomly selected pregnancy lasts less
than 247 days is approximately: ______
(Round to four decimal places as needed.)
Interpret this probability. Select the correct choice below and
fill in the answer box...

Suppose the lengths of the pregnancies of a certain animal are
approximately normally distributed with mean mu equals 194 days and
standard deviation sigma equals 13 days. Complete parts (a)
through (f) below. (a) What is the probability that a randomly
selected pregnancy lasts less than 190 days? The probability that
a randomly selected pregnancy lasts less than 190 days is
approximately . 3783. (Round to four decimal places as needed.)
Interpret this probability. Select the correct choice below and...

Suppose the lengths of the pregnancies of a certain animal are
approximately normally distributed with mean mu = 283 days and
standard deviation sigma = 29 days. (a) What is the probability
that a randomly selected pregnancy lasts less than 273 days? The
probability that a randomly selected pregnancy lasts less than 273
days is approximately (Round to four decimal places as needed.) (b)
What is the probability that a random sample of 11 pregnancies has
a mean gestation period...

Suppose the lengths of the pregnancies of a certain animal are
approximately normally distributed with mean mu equals 285
days
and standard deviation sigma equals 28 days. What is the
probability a random sample of size 20 will have a mean gestation
period within 10 days of the mean?
The probability that a random sample of size 20 will have a mean
gestation period within 10 days of the mean is _____ (Round to
four decimal places as needed.)

Suppose the lengths of the pregnancies of a certain animal are
approximately normally distributed with mean
mu equals 135 daysμ=135 days
and standard deviation
sigma equals 14 daysσ=14 days.
Complete parts (a) through (f) below.Click here to view the
standard normal distribution table (page 1).
LOADING...
Click here to view the standard normal distribution table (page
2).
LOADING...
(a) What is the probability that a randomly selected pregnancy
lasts less than
130130
days?The probability that a randomly selected pregnancy lasts...

Suppose the lengths of the pregnancies of a certain animal are
approximately normally distributed with mean
mu equals μ=197 daysand standard deviation sigma equals sσ=14
days. Complete parts (a) through (f) below.
(a) What is the probability that a randomly selected pregnancy
lasts less than 192 days?
The probability that a randomly selected pregnancy lasts less
than 192 days is approximately 0.3604 (Round to
four decimal places as needed.)
Interpret this probability. Select the correct choice below and
fill in...

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