Please help meeeeeee. image attached

When Noah goes bowling, his scores are normally distributed with a mean of 150 and a standard deviation of 12. What is the proba

Zarrin [17]

Answer:

The probability of his score being between 135 and 167 is 0.8151 or (0.8151*100=81.51%)

Step-by-step explanation:

Given that:

Mean = μ = 150

SD = σ = 12

Let x1 be the first data point and x2 the second data point

We have to find the z-scores for both data points

x1 = 135

x2 = 167

So,

$z_1 = \frac{x_1-mean}{SD}\\= \frac{135-150}{12}\\=\frac{-15}{12}\\=-1.25$

And

$z_2 = \frac{x_2-mean}{SD}\\z_2 =\frac{167-150}{12}\\=\frac{17}{12}\\= {1.416}$

We have to find area to the left of both points then their difference to find the probability.

So,

Area to the left of z1 = 0.1056

Area to the left of z2 = 0.9207

Probability to score between 135 and 167 = z2-z1 = 0.9027-0.1056 = 0.8151

Hence,

The probability of his score being between 135 and 167 is 0.8151 or (0.8151*100=81.51%)

3 0

2 years ago

A group of college students developed a solar-powered car and entered it in a race. The car travels at a constant speed of 100 m

nekit [7.7K]

Answer:

A) Distance time graph

B) d(t) = 25t

C) The expression shows the distance more clearly.

Step-by-step explanation:

A) A distance time graph as seen in the attachment provides a representation of the distance travelled.

We are told the car travels at a constant speed of 100 meters per 4 seconds. Which means that 100 m for each 4 hours. So, for 200m, it's 8 hours like seen in the graph and for 300m,it's 12 hours as seen in the graph.

B) And expression for the distance is;

d = vt

Where;

d is distance in metres

v is speed in m/s and t is time

We are told that the car travels at a constant speed of 100 meters per 4 seconds.

Thus, v = 100/4 = 25 m/s

Distance travelled over time is;

d(t) = 25t

C) Looking at both A and B above, it's obvious that the expression of the distance shows a more clearer way of getting the distance because once we know the time travelled, we will just plug it into the equation and get the distance. Whereas, for the representation form, one will need to longer graphs if the time spent is very long.