What are the last non-zero digits of 2015! ?

A set of exam scores is normally distributed and has a mean of 80.2 and a standard deviation of 11. What is the probability that

ZanzabumX [31]

Answer:

93.32% probability that a randomly selected score will be greater than 63.7.

Step-by-step explanation:

Problems of normally distributed samples are 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.

In this problem, we have that:

$\mu = 80.2, \sigma = 11$

What is the probability that a randomly selected score will be greater than 63.7.

This is 1 subtracted by the pvalue of Z when X = 63.7. So

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

$Z = \frac{63.7 - 80.2}{11}$

$Z = -1.5$

$Z = -1.5$ has a pvalue of 0.0668

1 - 0.0668 = 0.9332

93.32% probability that a randomly selected score will be greater than 63.7.

5 0

3 years ago

The number of accidents that occur at a busy intersection is Poisson distributed with a mean of 4.5 per week. Find the probabili

Mariana [72]

Answer:

a) 0.01111

b) 0.4679

c) 0.33747

Step-by-step explanation:

We are given the following in the question:

The number of accidents per week can be treated as a Poisson distribution.

Mean number of accidents per week = 4.5

$\lambda= 4.5$

Formula:

$P(X =k) = \displaystyle\frac{\lambda^k e^{-\lambda}}{k!}\\\\ \lambda \text{ is the mean of the distribution}$

a) No accidents occur in one week.

$P(x =0)\\\\= \displaystyle\frac{4.5^0 e^{-4.5}}{0!}= 0.01111$

b) 5 or more accidents occur in a week.

$P( x \geq 5) = 1-\displaystyle \sum P(x$

c) One accident occurs today.

The mean number of accidents per day is given by

$\lambda = \dfrac{4.5}{7} = 0.64$

$P(x =1)\\\\= \displaystyle\frac{0.64^1 e^{-0.64}}{1!}= 0.33747$

5 0

2 years ago

3. Leanne has a container divided into

Maurinko [17]

Answer:

22 containers but I didn't use a calculator!

3 0

3 years ago

Help it's due tomorrow!!!!

zepelin [54]

8a^2 + 3a^2

11a^2

since both have a^2 you only have to add the coefficients

5 0

3 years ago

The measures of 6 of the interior angles of a heptagon are: 120°, 150°, 135°, 170°, 90°, and 125°. What is the measure of the la

Lilit [14]

1. consider one angle of a (convex) heptagon. From that angle you can construct 7-3=4 diagonals. (-3 because we cannot create diagonals with the adjacent vertices and the angle itself )

2. 4 diagonals create 5 triangular regions. (check the picture)

3. So the sum of the measures of the interior angles of the heptagon is 180°*5=900°.

4. The measure of the remaining 7th interior angle is 900°-(120+150+135+170+90+125)°=110°.

5. The largest exterior angle is when the interior angle is the smallest.

6. The smallest interior angle is 90°, so the largest exterior angle is 180°-90°=90°

Answer: 90°

7 0

3 years ago