Answer:
Each hour he fly 107.76 miles
Step-by-step explanation:
we know that
Charles Lindbergh flew 3,610 miles in 33.5 hours
To find out the unit rate, divide the total distance by the total time
so
therefore
Each hour he fly 107.76 miles
The probability distribution for a random variable x is given in the table.
1 answer:
You might be interested in
Answer:
a) 0.277 = 27.7% probability that a randomly selected tyre lasts between 55,000 and 65,000 miles.
b) 0.348 = 34.8% probability that a randomly selected tyre lasts less than 48,000 miles.
c) 0.892 = 89.2% probability that a randomly selected tyre lasts at least 41,000 miles.
d) 0.778 = 77.8% probability that a randomly selected tyre has a lifetime that is within 10,000 miles of the mean
Step-by-step explanation:
Problems of normally distributed distributions 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 question, we have that:
Probabilities:
A) Between 55,000 and 65,000 miles
This is the pvalue of Z when X = 65000 subtracted by the pvalue of Z when X = 55000. So
X = 65000
has a pvalue of 0.954
X = 55000
has a pvalue of 0.677
0.954 - 0.677 = 0.277
0.277 = 27.7% probability that a randomly selected tyre lasts between 55,000 and 65,000 miles.
B) Less than 48,000 miles
This is the pvalue of Z when X = 48000. So
has a pvalue of 0.348
0.348 = 34.8% probability that a randomly selected tyre lasts less than 48,000 miles.
C) At least 41,000 miles
This is 1 subtracted by the pvalue of Z when X = 41,000. So
has a pvalue of 0.108
1 - 0.108 = 0.892
0.892 = 89.2% probability that a randomly selected tyre lasts at least 41,000 miles.
D) A lifetime that is within 10,000 miles of the mean
This is the pvalue of Z when X = 51200 + 10000 = 61200 subtracted by the pvalue of Z when X = 51200 - 10000 = 412000. So
X = 61200
has a pvalue of 0.889
X = 41200
has a pvalue of 0.111
0.889 - 0.111 = 0.778
0.778 = 77.8% probability that a randomly selected tyre has a lifetime that is within 10,000 miles of the mean