Steel rods are manufactured with a mean length of 29 centimeter (cm). Because of variability in the manufacturing process, the l
1 answer:
Solution :
Given data :
The mean length of the steel rod = 29 centimeter (cm)
The standard deviation of a normally distributed lengths of rods = 0.07 centimeter (cm)
a). We are required to find the proportion of rod that have a length of less than 28.9 centimeter (cm).
Therefore, P(x < 28.9) = P(z < (28.9-29) / 0.07)
= P(z < -1.42)
= 0.0778
b). Any rods which is shorter than cm or longer than
cm that re discarded.
Therefore,
P (x < 24.84 or 25.16 < x) = P( -59.42 < z or -54.85)
= 1.052
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Answer:
The expected value for a student to spend on lunch each day = $5.18
Step-by-step explanation:
Given the data:
Number of students______$ spent
2 students______________$10
1 student________________$8
12 students______________$6
23 students______________$5
8 students_______________$4
4 students_______________$3
Sample size, n = 50.
Let's first find the value on each amount spent with the formula:
Therefore,
For $10:
For $8:
l
For $6:
For $5:
For $4:
For $3:
To find the expected value a student spends on lunch each day, let's add all the values together.
Expected value =
$0.4 + $0.16 + 1.44 +$2.3 + $0.64 + $0.24
= $5.18
Therefore, the expected value for a student to spend on lunch each day is $5.18