Given:
To prove:
Proof:
In ,
because Given
because Given
because Reflexive property.
Corresponding sides of both triangles are congruent. So,
because SSS
Therefore, the required missing values are: Given, , Reflexive property ,
, SSS respectively.
The z-score of the speed value gives the measure of dispersion of the from
the mean observed speed.
The probability that the speed of a car is between 63 km/h and 75 km/h is
<u>0.273</u>.
The given parameters are;
The mean of the speed of cars on the highway, = 60 km/h
The standard deviation of the cars on the highway, σ = 5 km/h
Required:
The probability that the speed of a car is between 63 km/h and 75 km/h
Solution;
The z-score for a speed of 63 km/h is given as follows;
Which gives;
From the z-score table, we have;
P(x < 63) = 0.7257
The z-score for a speed of 75 km/h is given as follows;
Which gives, P(x < 75) = 0.9987
The probability that the speed of a car is between 63 km/h and 75 km/h is therefore;
P(63 < x < 75) = P(x < 75) - P(x < 63) = 0.9987 - 0.7257 = 0.273
The probability that the speed of a car is between 63 km/h and 75 km/h is
<u>0.273</u>.
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By considering the given information, we have
Null hypothesis :
Alternative hypothesis :
Since the alternative hypothesis is two-tailed , so the test is a two-tailed test.
Given : Sample size : n= 20, since sample size is less than 30 so the test applied is a t-test.
;
Test statistic :
i.e.
Degree of freedom : n-1 = 20-1=19
Significance level = 0.01
For two tailed, Significance level
By using the t-distribution table, the critical value of t =
Since , the observed t-value (7.25) is greater than the critical value (2.861) .
So we reject the null hypothesis, it means we have enough evidence to support the alternative hypothesis.
We conclude that there is some significance difference between the mean score for sober women and 35.0.