What is 5(h-4)=8?????????????????????
1 answer:
You might be interested in
Answer:
0.0326 = 3.26% probability that a randomly selected thermometer reads between −2.23 and −1.69.
The sketch is drawn at the end.
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Mean of 0°C and a standard deviation of 1.00°C.
This means that
Find the probability that a randomly selected thermometer reads between −2.23 and −1.69
This is the p-value of Z when X = -1.69 subtracted by the p-value of Z when X = -2.23.
X = -1.69
has a p-value of 0.0455
X = -2.23
has a p-value of 0.0129
0.0455 - 0.0129 = 0.0326
0.0326 = 3.26% probability that a randomly selected thermometer reads between −2.23 and −1.69.
Sketch:
Answer:
a) Null and alternative hypotheses are:
: mu=183 days
: mu>183 days
b) If the true mean is 190 days, Type II error can be made.
Step-by-step explanation:
Let mu be the mean life of the batteries of the company when it is used in a wireless mouse
Null and alternative hypotheses are:
: mu=183 days
: mu>183 days
Type II error happens if we fail to reject the null hypothesis, when actually the alternative hypothesis is true.
That is if we conclude that mean life of the batteries of the company when it is used in a wireless mouse is at most 183 days, but actually mean life is 190 hours, we make a Type II error.