Question

Question 1 Suppose that an experimenti setup to roll a fair 2. Suppose this experiment is done 25 times. While showing work and stating methods for calcula setup to roll a fair six-sided die A Success is recorded if the die shows a lor using a calculator, be sure to state the distribution) where work cannot be shown, find: (a) The probability that we have at most successes (5 points)

Answer 1

Question 1:

Binomial Distribution

n = 25

p = 2/6 = 0.3333

q = 1 - p = 0.6667

P(X$\leq$7) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7)

$P(X=0)=\binom{25}{0}\times 0.6667^{25}\times 0.3333^{0}=1\times 0.00003960 \times1 =0.00003960$

$P(X=1)=\binom{25}{1}\times 0.6667^{24}\times 0.3333^{1}=25\times 0.00005947 \times 0.3333=0.0004956$

$P(X=2)=\binom{25}{2}\times 0.6667^{23}\times 0.3333^{2}=300\times 0.00008921\times 0.1189=0.0030$

$P(X=3)=\binom{25}{3}\times 0.6667^{22}\times 0.3333^{3}=2300\times 0.0001338\times 0.0370=0.0114$

$P(X=4)=\binom{25}{4}\times 0.6667^{21}\times 0.3333^{4}=12650\times 0.0002007\times 0.0123=0.0313$

$P(X=5)=\binom{25}{5}\times 0.6667^{20}\times 0.3333^{5}=53130\times 0.0003010\times 0.0041=0.0658$

$P(X=6)=\binom{25}{6}\times 0.6667^{19}\times 0.3333^{6}=177100\times 0.0004515\times 0.0014=0.1096$

$P(X=7)=\binom{25}{7}\times 0.6667^{18}\times 0.3333^{7}=480700\times 0.0006772\times 0.0005=0.1488$

So,

P(X$\leq$7) = 0.3704

So,

Answer is:

0.3704

(74)

H0: Null Hypothesis: $\mu \geq$50000 (Average is at least 50,000) (Claim)

HA: Alternative Hypothesis: < 50,000

SE = $\sigma$/$\sqrt{n}$

= 8000/$\sqrt{28}$

= 1511.8579

Test Statistic is given by:

t = (46500 - 50000)/1511.8579

= - 2.3150

ndf = n - 1 28 - 1 = 27

$\alpha$ = 0.05

From Table, critical value of t = - 1.7033

Since calculated value of t = - 2.3150 is less than critical value of t = - 1.7033, the difference is significant. Reject null hypothesis.

Conclusion:
The data do not support the claim that tires averages at least 50,000 miles.