John watered

A professor at a local university noted that the exam grades of her students were normally distributed with a mean of 73 and a s

Umnica [9.8K]

Answer:

88.51 is the minimum score needed to receive a grade of A.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 73

Standard Deviation, σ = 11

We are given that the distribution of exam grades is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find the value of x such that the probability is 0.0793.

$P( X \geq x) = P( z \geq \displaystyle\frac{x - 73}{11})=0.0793$

$= 1 -P( z \leq \displaystyle\frac{x - 73}{11})=0.0793$

$=P( z \leq \displaystyle\frac{x - 73}{11})=0.9207$

Calculation the value from standard normal z table, we have,

$P(z \leq 1.410) = 0.9207$

$\displaystyle\frac{x - 73}{11} = 1.410\\x =88.51$

Hence, 88.51 is the minimum score needed to receive a grade of A.

3 0

3 years ago

A furniture store salesman had $15,200 in total monthly sales last month. He made $912 in commission from those sales.

tatyana61 [14]

We need to find out what percentage $912 is of $15 200. The answer is 6%, if I'm understanding the question correctly.

8 0

3 years ago

Read 2 more answers

It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute

Molodets [167]

Answer:

10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Mildly obese

Normally distributed with mean 375 minutes and standard deviation 68 minutes. So $\mu = 375, \sigma = 68$

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?

So $n = 6, s = \frac{68}{\sqrt{6}} = 27.76$

This probability is 1 subtracted by the pvalue of Z when X = 410.

$Z = \frac{X - \mu}{s}$

$Z = \frac{410 - 375}{27.76}$

$Z = 1.26$

$Z = 1.26$ has a pvalue of 0.8962.

So there is a 1-0.8962 = 0.1038 = 10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

Lean

Normally distributed with mean 522 minutes and standard deviation 106 minutes. So $\mu = 522, \sigma = 106$

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes?

So $n = 6, s = \frac{106}{\sqrt{6}} = 43.27$

This probability is 1 subtracted by the pvalue of Z when X = 410.

$Z = \frac{X - \mu}{s}$

$Z = \frac{410 - 523}{43.27}$

$Z = -2.61$

$Z = -2.61$ has a pvalue of 0.0045.

So there is a 1-0.0045 = 0.9955 = 99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

6 0

3 years ago

20 POINTS NO CAP PLEASE HELP ME NEED RIGHT ANSWER

jeka57 [31]

Answer:

yaaa

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Im stuck on this question, plz help

denis-greek [22]

Answer:

Kindly check explanation

Step-by-step explanation:

Given the following :

40 metres covered in 6 seconds

Max speed attained after 6seconds

75 meters covered after 11 seconds

Average Speed for first 40 meters :

Speed = distance / time

Speed = 40m / 6s

Speed = 6.67m/s

To obtain the maximum speed :

Next (75 - 40) meters = 35 meters was covered in (11 - 6)seconds = 5 seconds

Speed at this point is maximum :

Hence, maximum speed = (35m / 5s) = 7m/s

Suppose, Manuel runs for an additional z seconds after reaching max speed :

Distance from starting line 6+z seconds after race started?

Distance after 6 seconds = 40 metres

Distance after z seconds = 7 * z

Total distance = (40 + 7z)

What is Manuel's distance from the starting line x seconds after the race started (provided x≥6x)?

Distance for first 6 seconds = 40 meters + distance covered after 6 seconds = (7 * (x-6))

40 + 7(x - 6)

3 0

3 years ago