Question

**1. For a standard normal distribution,
find: **

P(z > 2.32)

Keep four decimal places.

**2. For a standard normal distribution,
find:**

P(-0.9 < z < 0.95)

**3. For a standard normal distribution,
given:**

P(z < c) = 0.7622

Find c.

**4. For a standard normal distribution,
find:**

P(z > c) = 0.1753

Find c

**5****.** Assume**that the readings at freezing on a batch of thermometers
are normally distributed with a mean of 0°C and a standard
deviation of 1.00°C. A single thermometer is randomly selected and
tested. Find the probability of obtaining a reading less than
-0.864°C.**

P(Z<−0.864)=P(Z<-0.864)=

**6.** **Assume** **that the
readings at freezing on a batch of thermometers are normally
distributed with a mean of 0°C and a standard deviation of 1.00°C.
A single thermometer is randomly selected and tested. Find the
probability of obtaining a reading between -0.495°C and
1.88°C.**

P(−0.495<Z<1.88)=P(-0.495<Z<1.88)=

**7.** **About % of the area under the curve
of the standard normal distribution is outside the interval
z=[−0.91,0.91]z=[-0.91,0.91] (or beyond 0.91 standard deviations of
the mean)**.

**8. Assume that the readings at freezing on a batch of
thermometers are normally distributed with a mean of 0°C and a
standard deviation of 1.00°C. A single thermometer is randomly
selected and tested. Find P_{11}, the
11-percentile. Round to 3 decimal places. This is the temperature
reading separating the bottom 11% from the top 89%.**

**9. z = -3 is what percentile?**

percentile

State your answer to the nearest tenth of a percent. Don't include
letters like "st", "nd", or "th" in your answer.

Answer #1

1:

The required probability, using z table, is

2;

The required probability, using z table, is

3:

Using z table we have

That is area left to z-score 0.71 is approximately 0.7622.

**Hence, c = 0.71**

4:

Using given equation we have

Using z table we have

That is area left to z-score 0.93 is approximately 0.8247.

**Hence, c = 0.93**

This is a multi-part question.
First, find the following: For a standard normal distribution,
find: P(-1.17 < z < 0.02)+__________
Second, resolve this: Assume that the readings at freezing on a
batch of thermometers are normally distributed with a mean of 0°C
and a standard deviation of 1.00°C. A single thermometer is
randomly selected and tested. Find the probability of obtaining a
reading less than -0.08°C.
P(Z<−0.08)= ____________
Third, resolve this: Assume that the readings at freezing on a
batch...

A. For a standard normal distribution, find:
P(-1.14 < z < -0.41)
B. For a standard normal distribution, given:
P(z < c) = 0.0278
Find c.
C. For a standard normal distribution, find:
P(z > c) = 0.4907
Find c.
D. Assume that z-scores are normally distributed with a mean
of 0 and a standard deviation of 1.
IfP(0<z<a)=0.4686P(0<z<a) = 0.4686
find a.
E. Assume that the readings at freezing on a bundle of
thermometers are normally distributed with a...

3.Assume that the readings at freezing on a batch of
thermometers are normally distributed with a mean of 0°C and a
standard deviation of 1.00°C. A single thermometer is randomly
selected and tested. Find the probability of obtaining a reading
greater than 1.865°C. P(Z>1.865)=P(Z>1.865)= (Round to four
decimal places)
4.Assume that the readings at freezing on a batch of
thermometers are normally distributed with a mean of 0°C and a
standard deviation of 1.00°C. A single thermometer is randomly
selected...

Assume that the readings at freezing on a batch of thermometers
are normally distributed with a mean of 0°C and a standard
deviation of 1.00°C. A single thermometer is randomly selected and
tested.
Find the probability of obtaining a reading between -2.75°C and
0°C. P ( − 2.75 < Z < 0 ) =

A.) Let XX represent the full height of a certain species of
tree. Assume that XX has a normal probability distribution with a
mean of 114.5 ft and a standard deviation of 7.8 ft.
A tree of this type grows in my backyard, and it stands 98.1 feet
tall. Find the probability that the height of a randomly selected
tree is as tall as mine or shorter.
P(X<98.1)P(X<98.1) =
My neighbor also has a tree of this type growing in...

Assume that the readings at freezing on a bundle of thermometers
are normally distributed with a mean of 0°C and a standard
deviation of 1.00°C. A single thermometer is randomly selected and
tested. Find the probability of obtaining a reading less than
-1.503°C.
P(Z<−1.503)=P(Z<-1.503)=

Assume that the readings at freezing on a bundle of thermometers
are normally distributed with a mean of 0°C and a standard
deviation of 1.00°C. A single thermometer is randomly selected and
tested. Find the probability of obtaining a reading between
-1.404°C and 2.955°C.
P(−1.404<Z<2.955)=

Assume that the readings at freezing on a bundle of thermometers
are normally distributed with a mean of 0°C and a standard
deviation of 1.00°C. A single thermometer is randomly selected and
tested. Find the probability of obtaining a reading between 0.244°C
and 0.251°C.
P(0.244<Z<0.251)

Assume that the readings at freezing on a batch of thermometers
are normally distributed with a mean of 0°C and a standard
deviation of 1.00°C. A single thermometer is randomly selected and
tested. Find P99, the 99-percentile. This is
the temperature reading separating the bottom 99% from the top
1%.

Assume that the readings at freezing on a bundle of thermometers
are normally distributed with a mean of 0°C and a standard
deviation of 1.00°C. A single thermometer is randomly selected and
tested. Find the probability of obtaining a reading between
-0.276°C and 1.304°C.
P(−0.276<Z<1.304)=P(-0.276<Z<1.304)=

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