A particular bulb has lifetimes of 1500 hours with a standard deviation of 6 hours. Find a range in which it can be guaranteed that 93.75% of the lifetimes of this brand of car lie.
(1489 and 1512)
(1482 and 1518)
(1476 and 1524)
(1470 and 1530)
Mean = 1500
Standard deviation = 6
Z value at 93.75% confidence interval = 1.534
"Mean\\pm Z \\frac{s}{\\sqrt{n}}"
"1500\\pm 1.534 \\frac{6}{\\sqrt{1}}"
"1500\\pm 1.534 \\times6"
(1489 and 1512)
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