Another way to write 300 + 70 + 5/10 + 8/100
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Answer:
a) 34.46% of 10-year-old boys is tall enough to ride this coaster.
b) 78.81% of 10-year-old boys is tall enough to ride this coaster
c) 44.35% of 10-year-old boys is tall enough to ride the coaster in part b but not tall enough to ride the coaster in part a
Step-by-step explanation:
When the distribution is normal, we use 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 question, we have that:
a. What proportion of 10-year-old boys is tall enough to ride the coaster?
This is 1 subtracted by the pvalue of Z when X = 56.
So
has a pvalue of 0.6554
1 - 0.6554 = 0.3446
34.46% of 10-year-old boys is tall enough to ride this coaster.
b. A smaller coaster has a height requirement of 50 inches to ride. What proportion of 10-year-old boys is tall enough to ride this coaster?
This is 1 subtracted by the pvalue of Z when X = 50.
has a pvalue of 0.2119
1 - 0.2119 = 0.7881
78.81% of 10-year-old boys is tall enough to ride this coaster.
c. What proportion of 10-year-old boys is tall enough to ride the coaster in part b but not tall enough to ride the coaster in part a?
Between 50 and 56 inches, which is the pvalue of Z when X = 56 subtracted by the pvalue of Z when X = 50.
From a), when X = 56, Z has a pvalue of 0.6554
From b), when X = 50, Z has a pvalue of 0.2119
0.6554 - 0.2119 = 0.4435
44.35% of 10-year-old boys is tall enough to ride the coaster in part b but not tall enough to ride the coaster in part a
Mean of the distribution = u = 222
Standard Deviation = s = 16
We have to find the probability that a value lies between 190 and 230.
First we need to convert these data values to z score.
For x = 190,
For x = 230
So, we have to find the percentage of values lying between z score of -2 and 0.5
P( -2 < z < 0.5) = P(0.5) - P(-2)
From standard z table, we can find and use these values.
P(-2 < x < 0.5 ) = 0.6915 - 0.0228 = 0.6687
Thus, there is 0.6887 probability that the data value will lie between 190 and 230 for the given distribution.
Standard Deviation = s = 16
We have to find the probability that a value lies between 190 and 230.
First we need to convert these data values to z score.
For x = 190,
For x = 230
So, we have to find the percentage of values lying between z score of -2 and 0.5
P( -2 < z < 0.5) = P(0.5) - P(-2)
From standard z table, we can find and use these values.
P(-2 < x < 0.5 ) = 0.6915 - 0.0228 = 0.6687
Thus, there is 0.6887 probability that the data value will lie between 190 and 230 for the given distribution.