This relationship is given by the equation, v = gt. In this equation, g is the acceleration due to gravity, t is the number of s
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Answer:
(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.
(b) The fraction of the calls last more than 5.30 minutes is 0.1271.
(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.
(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.
(e) The time is 5.65 minutes.
Step-by-step explanation:
We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.
Let X = <u><em>the length of the calls, in minutes.</em></u>
So, X ~ Normal()
The z-score probability distribution for the normal distribution is given by;
Z = ~ N(0,1)
where, = population mean time = 4.5 minutes
= standard deviation = 0.7 minutes
(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X 4.50 min)
P(X < 5.30 min) = P( <
) = P(Z < 1.14) = 0.8729
P(X 4.50 min) = P(
) = P(Z
0) = 0.50
The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.
Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = <u>0.3729</u>.
(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)
P(X > 5.30 min) = P( >
) = P(Z > 1.14) = 1 - P(Z
1.14)
= 1 - 0.8729 = <u>0.1271</u>
The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.
(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X 5.30 min)
P(X < 6.00 min) = P( <
) = P(Z < 2.14) = 0.9838
P(X 5.30 min) = P(
) = P(Z
1.14) = 0.8729
The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.
Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = <u>0.1109</u>.
(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X 4.00 min)
P(X < 6.00 min) = P( <
) = P(Z < 2.14) = 0.9838
P(X 4.00 min) = P(
) = P(Z
-0.71) = 1 - P(Z < 0.71)
= 1 - 0.7612 = 0.2388
The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.
Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = <u>0.745</u>.
(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;
P(X > x) = 0.05 {where x is the required time}
P( >
) = 0.05
P(Z > ) = 0.05
Now, in the z table the critical value of x which represents the top 5% of the area is given as 1.645, that is;
x = 4.5 + 1.15 = 5.65 minutes.
SO, the time is 5.65 minutes.
Answer:
a) Mean blood pressure for people in China.
b) 38.21% probability that a person in China has blood pressure of 135 mmHg or more.
c) 71.30% probability that a person in China has blood pressure of 141 mmHg or less.
d) 8.51% probability that a person in China has blood pressure between 120 and 125 mmHg.
e) Since Z when X = 135 is less than two standard deviations from the mean, it is not unusual for a person in China to have a blood pressure of 135 mmHg
f) 157.44mmHg
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.
If X is two standard deviations from the mean or more, it is considered unusual.
In this question:
a.) State the random variable.
Mean blood pressure for people in China.
b.) Find the probability that a person in China has blood pressure of 135 mmHg or more.
This is 1 subtracted by the pvalue of Z when X = 135.
has a pvalue of 0.6179
1 - 0.6179 = 0.3821
38.21% probability that a person in China has blood pressure of 135 mmHg or more.
c.) Find the probability that a person in China has blood pressure of 141 mmHg or less.
This is the pvalue of Z when X = 141.
has a pvalue of 0.7140
71.30% probability that a person in China has blood pressure of 141 mmHg or less.
d.)Find the probability that a person in China has blood pressure between 120 and 125 mmHg.
This is the pvalue of Z when X = 125 subtracted by the pvalue of Z when X = 120. So
X = 125
has a pvalue of 0.4483
X = 120
has a pvalue of 0.3632
0.4483 - 0.3632 = 0.0851
8.51% probability that a person in China has blood pressure between 120 and 125 mmHg.
e.) Is it unusual for a person in China to have a blood pressure of 135 mmHg? Why or why not?
From b), when X = 135, Z = 0.3
Since Z when X = 135 is less than two standard deviations from the mean, it is not unusual for a person in China to have a blood pressure of 135 mmHg.
f.) What blood pressure do 90% of all people in China have less than?
This is the 90th percentile, which is X when Z has a pvalue of 0.28. So X when Z = 1.28. Then
X = 120
So
157.44mmHg