7.5 is already in decimal form.
Answer:
a) 48.21 %
b) 45.99 %
c) 20.88 %
d) 42.07 %
e) 50 %
Note: these values represent differences between z values and the mean
Step-by-step explanation:
The test to carry out is:
Null hypothesis H₀ is μ₀ = 30
The alternative hypothesis m ≠ 30
In which we already have the value of z for each case therefore we look directly the probability in z table and carefully take into account that we had been asked for differences from the mean (0.5)
a) z = 2.1 correspond to 0.9821 but mean value is ubicated at 0.5 then we subtract 0.9821 - 0.5 and get 0.4821 or 48.21 %
b) z = -1.75 P(m) = 0.0401 That implies the probability of m being from that point p to the end of the tail, the difference between this point and the mean so 0.5 - 0.0401 = 0.4599 or 45.99 %
c) z = -.55 P(m) = 0.2912 and this value for same reason as before is 0.5 - 0.2912 = 0.2088 or 20.88 %
d) z = 1.41 P(m) = 0.9207 0.9207 -0.5 0.4207 or 42.07 %
e) z = -5.3 P(m) = 0 meaning there is not such value in z table is too small to compute and difference to mean value will be 0.5
d) z= 1.41 P(m) =
Answer:
85.5
Step-by-step explanation:
because you have to add 1/2 and 90cents =90.5 then you have to subtract 5 away and your answer is 85.5
Answer:
'A' is true; theoretically, 50% of the data items reside between the first and third quartiles (40 and 67.5)
Step-by-step explanation:
Range is 84-28 which is 56
Median is 51
1.5 x Interquartile Range (IQR) = 1.5(67.5-40) which equals 41.25
Q1 is 40
Q1 - IQR = -1.25
Low outliers are below -1.25 - 41.25; there are not data items below -42
