Answer:
Explanation:
Using the formula for finding the Z score
Z = x-µ/σ
x is the sample size
µ is the sample mean
σ is the standard deviation
For percentage of the resistors will have resistance between 148 Ω and 152 Ω, or is calculated as shown
P(148≤x152) = Z(152-150/5) - Z(148-150/5)
P(148≤x152) = Z(0.4)-Z( - 0.4)
P(148≤x152) = 0.6554-0.3446
The Z values are from the normal distribution table.
P(148≤x152) = 0.3108
The percentage of resistor that will have between 148 and 152 ohms is 0.3108×100% = 31.08%
Similarly for resistances between 140 Ω and 160 Ω
P(140≤x160) = Z(160-150/5) - Z(140-150/5)
P(140≤x160) = Z(2.0)-Z( - 2.0)
P(140≤x160) = 0.9775-0.02275
The Z values are from the normal distribution table.
P(140≤x160) = 0.9547
The percentage of resistor that will have between 140 and 160ohms is 0.9547×100% = 95.47%
The percentage of the resistors will have resistance between 148 Ω and 152 Ω is 31.08%
The percentage of the resistors will have resistance between 140 Ω and 160 Ω is 95.47%
To solve for the probability we will use the standard score of the Z score, which is given by:
Z = (x - µ)/σ
where x is the sample size
µ is the sample mean = 150Ω
σ is the standard deviation = 5Ω
The probability of the resistors with resistance between 148 Ω and 152 Ω, will be:
P(148 ≤ 152) = Z((152-150)/5) - Z((148-150)/5)
P(148 ≤ 152) = Z(0.4)-Z( - 0.4)
P(148 ≤ 152) = 0.6554-0.3446
P(148 ≤ 152) = 0.3108
So, the percentage will be:
0.3108×100% = 31.08%
Similarly for resistances between 140 Ω and 160 Ω
P(140 ≤ 160) = Z((160-150)/5) - Z((140-150)/5)
P(140 ≤ 160) = Z(2.0)-Z( - 2.0)
P(140 ≤ 160) = 0.9775-0.02275
P(140 ≤ 160) = 0.9547
The percentage of resistor that will have between 140 and 160ohms is 0.9547×100% = 95.47%
Learn more about standard score:
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