Question

The owner of a meat market has an assistant who has determined
that the weights of roasts are normally distributed, with a mean of
3.2 pounds and standard deviation of 0.8 pounds. If a sample of 25
roasts yields a mean of 3.6 pounds, what is the *Z*-score
for this sample mean?

Group of answer choices

1. None of these choices.

2. −2.50

3. . 2.50

4. −0.50

Answer #1

Population mean, = 3.2 pounds

Population standard deviation, = 0.8 pounds

Sample size, n = 25

For sampling distribution of mean,

Mean, = = 3.2 pounds

Standard error, =

=

= 0.16

Here, = 3.6 pounds

Z score of sample mean = ( - )/

= (3.6 - 3.2)/0.16

= **2.5**

The owner of a fish market has an assistant who has determined
that the weights of catfish are normally distributed, with a mean
of 3.2 pounds and a standard deviation of 0.8 pound. If a sample of
64 fish yields a mean of 3.4 pounds, what is probability of
obtaining a sample mean this large or larger?
0.4987
0.0013
0.0001
0.0228

The owner of a fish market has an assistant who has determined
that the weights of catfish are normally distributed, with a mean
of 3.2 pounds and a standard deviation of 0.8 pound. If a sample of
16 fish is taken, what would the standard error of the mean weight
equal?
0.003
0.200
0.800
0.050

The owner of a fish market has an assistant who has determined
that the weights of catfish are normally distributed, with mean of
3.2 pounds and standard deviation of 0.8 pound. A sample of 4 fish
is taken. Below what value do 89.62% of the sample mean fall?
A
2.696
B
2.842
C
3.559
D
3.704

The owner of a fish market has an assistant who has determined
that the weights of catfish are normally distributed with a mean of
3.2 pounds and a standard deviation of .8 pounds. What is the
probability that a sample of 64 fish will have a sample mean
between 2.9 and 3.4 pounds?

1. The weights of a certain dog breed are approximately normally
distributed with a mean of ? = 46 pounds, and a standard deviation
of ? = 7 pounds.
A) A dog of this breed weighs 51 pounds. What is the dog's
z-score? Round your answer to the nearest hundredth as needed. z
=
B) A dog has a z-score of -0.8. What is the dog's weight? Round
your answer to the nearest tenth as needed. ____ pounds
C) A...

The distribution of weights of a sample of 500 toddlers is
symmetric and bell-shaped.
According to the Empirical Rule, what percent of the weights will
lie between plus or
minus three sigmas (standard deviations) from the mean?
Group of answer choices
68%
34%
99.7%
95%
None of these
If a sample of toddlers has an estimated mean weight of 52
pounds and a standard
deviation of 4 pounds, then 95% of the toddlers have weights
between which two
values?
If...

The owner of a computer repair shop has determined that their
daily revenue has mean $7200 and standard deviation $1200. The
daily revenue is normally distributed. a) What is the probability
that a randomly selected day will have a revenue of at most $7000?
b) The daily revenue for the next 30 days will be monitored. What
is the probability that the mean daily revenue for the next 30 days
will exceed $7500?

se z scores to compare the given values. Based on sample data,
newborn males have weights with a mean of 3258.2 g and a standard
deviation of 580.4 g. Newborn females have weights with a mean of
3093.6 g and a standard deviation of 643.6 g. Who has the weight
that is more extreme relative to the group from which they came: a
male who weighs 1700 g or a female who weighs 1700 g? Since the z
score for...

Use z scores to compare the given values. Based on sample data,
newborn males have weights with a mean of 3271.5 g and a standard
deviation of 663.6 g. Newborn females have weights with a mean of
3049.4 g and a standard deviation of 759.9 g. Who has the weight
that is more extreme relative to the group from which they came: a
male who weighs 1500 g or a female who weighs 1500 g? Since the z
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1. Find P(- 1.68 < Z < 1.68)
Group of answer choices
A. .9535
B. .9070
C. .0465
D. .4535
2. A study of homeowners in the 5th congressional district in
Maryland found that their annual
household incomes are normally distributed with a mean of $41,182
and a standard deviation of $11,990
(based on data from Nielsen Media Research).
What percentage of household incomes are greater than
$30,000?
Group of answer choices
A. 31.42%
B. 82.45%
C. 32.45%
D. 17.55%...

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