Question

1. the mean life of a battery is 50 hours with a standard deviation of 6 hours. the manufacturer advertises that they will replace all batteries that last less than 38 hours. if 50,000 batteries were produced, how many would they expect to replace?

normalcdf ()=

1. a grading scale is set up for 1000 students test scores. it is assumed that the scores are normally distributed with a mean score of 75 and a standard deviation of 15

a) how many students will have scores higher than 83?
normalcdf ()=
b) how many students will have scores below 50?
normalcdf ()=
c) how many students will have scores between 48 and 75?
normalcdf ()=

Explanation:

Step1: Calculate z - score for battery problem

The z - score formula is \(z=\frac{x-\mu}{\sigma}\). For the battery problem, \(\mu = 50\), \(\sigma=6\), and \(x = 38\).

$$z=\frac{38 - 50}{6}=\frac{-12}{6}=- 2$$

We use the normalcdf function. In the TI - 84 calculator, normalcdf\((-\infty,38,50,6)\) is equivalent to normalcdf\((-1000,38,50,6)\) (where \(-1000\) is used to approximate \(-\infty\)).

$$normalcdf(-1000,38,50,6)$$

Using the properties of the normal distribution, the probability \(P(X < 38)\) corresponding to \(z=-2\) is \(0.0228\) (from the standard normal table or calculator).
The number of batteries to replace is \(n = 50000\times0.0228\)

Step2: Calculate for part (a) of the test - score problem

For part (a), \(\mu = 75\), \(\sigma = 15\), \(x = 83\)

$$z=\frac{83 - 75}{15}=\frac{8}{15}\approx0.533$$

We use normalcdf\((83,\infty,75,15)\) which is equivalent to normalcdf\((83,1000,75,15)\)

$$normalcdf(83,1000,75,15)$$

Using the calculator, the probability \(P(X>83)=1 - P(X\leq83)\). From the calculator, \(P(X\leq83)\approx0.697\), so \(P(X > 83)=1 - 0.697 = 0.303\)
The number of students is \(n=1000\times0.303\)

Step3: Calculate for part (b) of the test - score problem

For part (b), \(\mu = 75\), \(\sigma = 15\), \(x = 50\)

$$z=\frac{50 - 75}{15}=\frac{-25}{15}\approx - 1.67$$

We use normalcdf\((-\infty,50,75,15)\) which is equivalent to normalcdf\((-1000,50,75,15)\)

$$normalcdf(-1000,50,75,15)$$

Using the calculator, the probability \(P(X < 50)\approx0.0475\)
The number of students is \(n = 1000\times0.0475\)

Step4: Calculate for part (c) of the test - score problem

For part (c), \(\mu = 75\), \(\sigma = 15\), \(x_1 = 48\), \(x_2 = 75\)

$$z_1=\frac{48 - 75}{15}=\frac{-27}{15}=-1.8$$

$$z_2=\frac{75 - 75}{15}=0$$

We use normalcdf\((48,75,75,15)\)

$$normalcdf(48,75,75,15)$$

Using the calculator, \(P(48<X<75)\). Since \(P(X < 75)=0.5\) and \(P(X < 48)\approx0.0359\), \(P(48<X<75)=0.5 - 0.0359 = 0.4641\)
The number of students is \(n = 1000\times0.4641\)

Answer:

• Battery problem: normalcdf\((-1000,38,50,6)=0.0228\), number of batteries to replace \(=50000\times0.0228 = 1140\)
• Part (a): normalcdf\((83,1000,75,15)=0.303\), number of students \(=1000\times0.303 = 303\)
• Part (b): normalcdf\((-1000,50,75,15)=0.0475\), number of students \(=1000\times0.0475 = 47.5\approx48\)
• Part (c): normalcdf\((48,75,75,15)=0.4641\), number of students \(=1000\times0.4641 = 464.1\approx464\)