Suppose the amounts of M&M's a machine dispenses into packages has a normal distribution with a mean of 25 M&M's and a s
tandard deviation of 1 M&M. What is the probability that there are more than 26 M&M's?
1 answer:
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Answer:
a) 96.64% probability that maximum speed is at most 50 km/h
b) 24.67% probability that maximum speed is at least 48 km/h
c) 86.64% probability that maximum speed differs from the mean value by at most 1.5 standard deviations
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 is the probability that maximum speed is at most 50 km/h?
This is the pvalue of Z when X = 50. So
has a pvalue of 0.9664
96.64% probability that maximum speed is at most 50 km/h.
B. What is the probability that maximum speed is at least 48 km/h?
This is 1 subtracted by the pvalue of Z when X = 48.
has a pvalue of 0.7533
1 - 0.7533 = 0.2467
24.67% probability that maximum speed is at least 48 km/h.
C. What is the probability that maximum speed differs from the mean value by at most 1.5 standard deviations?
Z = 1.5 has a pvalue of 0.9332
Z = -1.5 has a pvalue of 0.0668
0.9332 - 0.0668 = 0.8664
86.64% probability that maximum speed differs from the mean value by at most 1.5 standard deviations