Can any one help me with this problem
1 answer:
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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:
$576.80
Step-by-step explanation:
We have been given that Mr. Juárez opened a savings account with an initial deposit of $560 and will not make any additional deposits or withdrawals. The account earns 1% simple interest.
We are asked to find the total amount that Mr. Juárez will have in his account at the end of 3 years.
We will use simple interest formula to solve our given problem.
A = Final amount after t years,
P = Principal amount,
r = Annual interest rate in decimal form,
t = Time in years.
Let us convert 1% into decimal form,
1%=1/100=0.01
P=$560 and t=3
A=$560 (1+0.01(3))
A=$560 (1+0.03)
A= $560 (1.03)
A= $576.80
Therefore, Mr. Juárez will have $576. 80 in his account at the end of 3 years. Hope this helps!