Given :
Admission price , A = $15 .
Price of each ride , p = $3 .
Number of rides , r = 3 .
To Find :
Total money spend if he goes on 5 rides and for r number of rides .
Solution :
Let , total amount paid is P .
First we find the general equation :
Now , for r = 5 , total price is :
Hence , this is the required solution .
Answer:
a) 68.26% probability that a student scores between 350 and 550
b) A score of 638(or higher).
c) The 60th percentile of test scores is 475.3.
d) The middle 30% of the test scores is between 411.5 and 488.5.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
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:
a. What is the probability that a student scores between 350 and 550?
This is the pvalue of Z when X = 550 subtracted by the pvalue of Z when X = 350. So
X = 550
has a pvalue of 0.8413
X = 350
has a pvalue of 0.1587
0.8413 - 0.1587 = 0.6826
68.26% probability that a student scores between 350 and 550
b. If the upper 3% scholarship, what score must a student receive to get a scholarship?
100 - 3 = 97th percentile, which is X when Z has a pvalue of 0.97. So it is X when Z = 1.88
A score of 638(or higher).
c. Find the 60th percentile of the test scores.
X when Z has a pvalue of 0.60. So it is X when Z = 0.253
The 60th percentile of test scores is 475.3.
d. Find the middle 30% of the test scores.
50 - (30/2) = 35th percentile
50 + (30/2) = 65th percentile.
35th percentile:
X when Z has a pvalue of 0.35. So X when Z = -0.385.
65th percentile:
X when Z has a pvalue of 0.35. So X when Z = 0.385.
The middle 30% of the test scores is between 411.5 and 488.5.
