Answer:
88.88% probability that it endures for less than a year and a half
Step-by-step explanation:
Problems of normally distributed samples can be 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:
The next career begins on Monday; what is the likelihood that it endures for less than a year and a half?
One year has 52.14 weeks. So a year and a half has 1.5*52.14 = 78.21 weeks.
So this probability is the pvalue of Z when X = 78.21.
has a pvalue of 0.8888
88.88% probability that it endures for less than a year and a half
Answer:
Option E) 61.6
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 100 bushels per acre
Standard Deviation, σ = 30 bushels per acre
We assume that the distribution of yield is a bell shaped distribution that is a normal distribution.
Formula:
P(X>x) = 0.90
We have to find the value of x such that the probability is 0.90
P(X > x)
Calculation the value from standard normal table, we have,
Hence, the yield of 61.6 bushels per acre or more would save the seed.
Answer:
Time(min)
2
6
8
15
Distance(km)
6
18
24
45
Step-by-step explanation:
the time is 1/3rd of the distance. it's a 1:3 ratio