All categories

2 years ago

Class A has 15 pupils and class B has 29 pupils.

Mathematics

1 answer:

Vaselesa [24]2 years ago

6 0

The mean score (rounded to 2 DP) in the Math's test for class B is 86.86.

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

Class A has 15 pupils and class B has 29 pupils. The mean score for class A is 55. Hence:

Mean score of A = (sum of all scores) / 15

55 = (sum of all scores) / 15

sum of all scores in A = 825

The mean score for both classes is 76. Hence:

[(sum of all scores in A) + (sum of all scores in B)]/(total number of pupils) = mean score of both classes

[825 + (sum of all scores in B)]/(15 + 29) = 76

sum of all scores in B = 2519

Mean score of B = (2519) / 29 = 86.86

The mean score (rounded to 2 DP) in the Math's test for class B is 86.86.

Find out more on equation at: brainly.com/question/13763238

#SPJ1

You might be interested in

Who know how to do this?

larisa86 [58]

Answer:

I think this is correct. I use demos calculator in 8th grade

6 0

3 years ago

16/14 as mixed number

SOVA2 [1]

This equals one and one eighth

7 0

3 years ago

The length of a car is √185 ft. Will this car fit in a garage that measure 10 ft long? If not that is the minimum length of a ga

Rina8888 [55]

10 is the square root of only 100, so I know that the square root of 185 must be longer.

The square root of 185 is about 13.6 , so the garage must be absolutely
not any shorter than 14-ft, and 15-ft or 16-ft would be even better.

4 0

4 years ago

The amount of time that people spend at Grover Hot Springs Spa is normally distributed with a mean of 63 minutes and a standard

PtichkaEL [24]

Answer:

a) $X \sim N(63,18)$

Where $\mu=63$ and $\sigma=18$

b) $P(X>90)=P(\frac{X-\mu}{\sigma}>\frac{90-\mu}{\sigma})=P(Z>\frac{90-63}{18})=P(Z>1.5)$

And we can find this probability using the complement rule and with the normal standard table or excel:

$P(Z>1.5)=1-P(Z$

c) $P(X$

$P(Z$

d) $P(60$

$P(-0.167$

e) IQR = 75.13-50.87=24.26

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent the amount of time that people spend at Grover Hot Springs of a population, and for this case we know the distribution for X is given by:

$X \sim N(63,18)$

Where $\mu=63$ and $\sigma=18$

Part b

We are interested on this probability

$P(X>90)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>90)=P(\frac{X-\mu}{\sigma}>\frac{90-\mu}{\sigma})=P(Z>\frac{90-63}{18})=P(Z>1.5)$

And we can find this probability using the complement rule and with the normal standard table or excel:

$P(Z>1.5)=1-P(Z$

Part c

We are interested on this probability

$P(X$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X$

And we can find this probability using the normal standard table or excel:

$P(Z$

Part d

If we apply this formula to our probability we got this:

$P(60$

And we can find this probability with this difference:

$P(-0.167$

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

$P(-0.167$

Part e

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.75$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.25 of the area on the left and 0.75 of the area on the right it's z=-0.674. On this case P(Z<-0.674)=0.25 and P(z>-0.674)=0.75

If we use condition (b) from previous we have this:

$P(X$