Answer:
4.05% probability that a randomly selected adult has an IQ greater than 123.4.
Step-by-step explanation:
Problems of normally distributed samples 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:
Probability that a randomly selected adult has an IQ greater than 123.4.
This is 1 subtracted by the pvalue of Z when X = 123.4. So
has a pvalue of 0.9595
1 - 0.9595 = 0.0405
4.05% probability that a randomly selected adult has an IQ greater than 123.4.
Answer: 1
/2 (
x
2 − 16
x + 6
)
Step-by-step explanation:
Ii is the correct answer and try using photomath it helps ;)
Actual area of the rectangular building = 5040 feet²
Step-by-step explanation:
Length of the rectangular building = 7 feet
Width of the rectangular building = 5 feet
Now, The building is made with a scale factor of 12
So, Actual length of the rectangular building = 84 feet
Actual width of the rectangular building = 60 feet
Now, Actual area of the rectangular building = Length × Width
⇒ Actual area of the rectangular building = 84 × 60
⇒ Actual area of the rectangular building = 5040 feet²
Answer: We reject the null hypothesis, and we use Normal distribution for the test.
Step-by-step explanation:
Since we have given that
We claim that
Null hypothesis : 
Alternate hypothesis : 
There is 5% level of significance.
So, the test statistic would be
Since alternate hypothesis is left tailed test.
So, p-value = P(z≤-2.31)=0.0401
And the P-value =0.0401 is less than the given level of significance i.e. 5% 0.05.
So, we reject the null hypothesis, and we use Normal distribution for the test.
